Integrand size = 40, antiderivative size = 289 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {(b B-3 a C) \text {arctanh}(\sin (c+d x))}{b^4 d}-\frac {a \left (2 a^4 b B-5 a^2 b^3 B+6 b^5 B-6 a^5 C+15 a^3 b^2 C-12 a b^4 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}-\frac {\left (a b B-3 a^2 C+2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac {a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {a^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:
(B*b-3*C*a)*arctanh(sin(d*x+c))/b^4/d-a*(2*B*a^4*b-5*B*a^2*b^3+6*B*b^5-6*C *a^5+15*C*a^3*b^2-12*C*a*b^4)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b) ^(1/2))/(a-b)^(5/2)/b^4/(a+b)^(5/2)/d-1/2*(B*a*b-3*C*a^2+2*C*b^2)*tan(d*x+ c)/b^3/(a^2-b^2)/d+1/2*a*(B*b-C*a)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/( a+b*sec(d*x+c))^2-1/2*a^2*(B*a^2*b-4*B*b^3-3*C*a^3+6*C*a*b^2)*tan(d*x+c)/b ^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))
Time = 7.03 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {a \left (2 a^4 b B-5 a^2 b^3 B+6 b^5 B-6 a^5 C+15 a^3 b^2 C-12 a b^4 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2} \left (-a^2+b^2\right )^2 d}+\frac {(-b B+3 a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d}+\frac {(b B-3 a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d}+\frac {C \sin \left (\frac {1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {C \sin \left (\frac {1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {a^2 b B \sin (c+d x)-a^3 C \sin (c+d x)}{2 b^2 (-a+b) (a+b) d (b+a \cos (c+d x))^2}+\frac {-2 a^4 b B \sin (c+d x)+5 a^2 b^3 B \sin (c+d x)+4 a^5 C \sin (c+d x)-7 a^3 b^2 C \sin (c+d x)}{2 b^3 (-a+b)^2 (a+b)^2 d (b+a \cos (c+d x))} \] Input:
Integrate[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[ c + d*x])^3,x]
Output:
(a*(2*a^4*b*B - 5*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 15*a^3*b^2*C - 12*a*b^4* C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b ^2]*(-a^2 + b^2)^2*d) + ((-(b*B) + 3*a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(b^4*d) + ((b*B - 3*a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2] ])/(b^4*d) + (C*Sin[(c + d*x)/2])/(b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x) /2])) + (C*Sin[(c + d*x)/2])/(b^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (a^2*b*B*Sin[c + d*x] - a^3*C*Sin[c + d*x])/(2*b^2*(-a + b)*(a + b)*d*( b + a*Cos[c + d*x])^2) + (-2*a^4*b*B*Sin[c + d*x] + 5*a^2*b^3*B*Sin[c + d* x] + 4*a^5*C*Sin[c + d*x] - 7*a^3*b^2*C*Sin[c + d*x])/(2*b^3*(-a + b)^2*(a + b)^2*d*(b + a*Cos[c + d*x]))
Time = 2.12 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.13, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4560, 3042, 4517, 3042, 4578, 3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4517 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-\left (\left (-3 C a^2+b B a+2 b^2 C\right ) \sec ^2(c+d x)\right )-2 b (b B-a C) \sec (c+d x)+2 a (b B-a C)\right )}{(a+b \sec (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (3 C a^2-b B a-2 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 b (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a (b B-a C)\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4578 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (-b \left (a^2-b^2\right ) \left (-3 C a^2+b B a+2 b^2 C\right ) \sec ^2(c+d x)+\left (a^2-b^2\right ) \left (-3 C a^3+b B a^2+4 b^2 C a-2 b^3 B\right ) \sec (c+d x)+a b \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right )\right )}{a+b \sec (c+d x)}dx}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-b \left (a^2-b^2\right ) \left (-3 C a^2+b B a+2 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (a^2-b^2\right ) \left (-3 C a^3+b B a^2+4 b^2 C a-2 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a b \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sec (c+d x) \left (a \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right ) b^2+2 \left (a^2-b^2\right )^2 (b B-3 a C) \sec (c+d x) b\right )}{a+b \sec (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \left (-3 C a^3+b B a^2+6 b^2 C a-4 b^3 B\right ) b^2+2 \left (a^2-b^2\right )^2 (b B-3 a C) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (b B-3 a C) \int \sec (c+d x)dx-a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (b B-3 a C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (b B-3 a C) \text {arctanh}(\sin (c+d x))}{d}-a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (b B-3 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (b B-3 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (b B-3 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}+\frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}+\frac {\frac {\frac {\frac {2 \left (a^2-b^2\right )^2 (b B-3 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a \left (-6 a^5 C+2 a^4 b B+15 a^3 b^2 C-5 a^2 b^3 B-12 a b^4 C+6 b^5 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {\left (a^2-b^2\right ) \left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}\) |
Input:
Int[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d* x])^3,x]
Output:
(a*(b*B - a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sec[ c + d*x])^2) + (-((a^2*(a^2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*C)*Tan[c + d *x])/(b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))) + (((2*(a^2 - b^2)^2*(b*B - 3*a*C)*ArcTanh[Sin[c + d*x]])/d - (2*a*(2*a^4*b*B - 5*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 15*a^3*b^2*C - 12*a*b^4*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x) /2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d))/b - ((a^2 - b^2)*(a*b*B - 3*a^2*C + 2*b^2*C)*Tan[c + d*x])/d)/(b^2*(a^2 - b^2)))/(2*b*(a^2 - b^2))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*d^2*( A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[d/(b*(m + 1)*(a^2 - b^2)) Int[( a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*( n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) - d*B*(a^2 *(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f , A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[ n, 1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x ])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - Simp[1/(b^2*(m + 1)*(a^2 - b^ 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(m + 1)*((-a)*(b *B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 ]
Time = 1.23 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.31
| method | result | size |
| derivativedivides | \(\frac {-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}+\frac {2 a \left (\frac {\frac {\left (2 B \,a^{2} b -B a \,b^{2}-6 B \,b^{3}-4 a^{3} C +a^{2} b C +8 C a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (2 B \,a^{2} b +B a \,b^{2}-6 B \,b^{3}-4 a^{3} C -a^{2} b C +8 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (2 B \,a^{4} b -5 B \,a^{2} b^{3}+6 B \,b^{5}-6 a^{5} C +15 C \,a^{3} b^{2}-12 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\) | \(380\) |
| default | \(\frac {-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{4}}+\frac {2 a \left (\frac {\frac {\left (2 B \,a^{2} b -B a \,b^{2}-6 B \,b^{3}-4 a^{3} C +a^{2} b C +8 C a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (2 B \,a^{2} b +B a \,b^{2}-6 B \,b^{3}-4 a^{3} C -a^{2} b C +8 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (2 B \,a^{4} b -5 B \,a^{2} b^{3}+6 B \,b^{5}-6 a^{5} C +15 C \,a^{3} b^{2}-12 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}-\frac {C}{b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +3 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{4}}}{d}\) | \(380\) |
| risch | \(\text {Expression too large to display}\) | \(1691\) |
Input:
int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x,method =_RETURNVERBOSE)
Output:
1/d*(-C/b^3/(tan(1/2*d*x+1/2*c)+1)+(B*b-3*C*a)/b^4*ln(tan(1/2*d*x+1/2*c)+1 )+2*a/b^4*((1/2*(2*B*a^2*b-B*a*b^2-6*B*b^3-4*C*a^3+C*a^2*b+8*C*a*b^2)*a*b/ (a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*b*a*(2*B*a^2*b+B*a*b^2-6*B* b^3-4*C*a^3-C*a^2*b+8*C*a*b^2)/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c))/(tan(1/2* d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2-1/2*(2*B*a^4*b-5*B*a^2*b^3+6* B*b^5-6*C*a^5+15*C*a^3*b^2-12*C*a*b^4)/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^( 1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))-C/b^3/(tan(1/2 *d*x+1/2*c)-1)+1/b^4*(-B*b+3*C*a)*ln(tan(1/2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 1026 vs. \(2 (277) = 554\).
Time = 25.00 (sec) , antiderivative size = 2111, normalized size of antiderivative = 7.30 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
Output:
[-1/4*(((6*C*a^8 - 2*B*a^7*b - 15*C*a^6*b^2 + 5*B*a^5*b^3 + 12*C*a^4*b^4 - 6*B*a^3*b^5)*cos(d*x + c)^3 + 2*(6*C*a^7*b - 2*B*a^6*b^2 - 15*C*a^5*b^3 + 5*B*a^4*b^4 + 12*C*a^3*b^5 - 6*B*a^2*b^6)*cos(d*x + c)^2 + (6*C*a^6*b^2 - 2*B*a^5*b^3 - 15*C*a^4*b^4 + 5*B*a^3*b^5 + 12*C*a^2*b^6 - 6*B*a*b^7)*cos( d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2 )/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) + 2*((3*C*a^9 - B*a^8*b - 9*C*a^7*b^2 + 3*B*a^6*b^3 + 9*C*a^5*b^4 - 3*B*a^4*b^5 - 3*C*a^3*b^6 + B *a^2*b^7)*cos(d*x + c)^3 + 2*(3*C*a^8*b - B*a^7*b^2 - 9*C*a^6*b^3 + 3*B*a^ 5*b^4 + 9*C*a^4*b^5 - 3*B*a^3*b^6 - 3*C*a^2*b^7 + B*a*b^8)*cos(d*x + c)^2 + (3*C*a^7*b^2 - B*a^6*b^3 - 9*C*a^5*b^4 + 3*B*a^4*b^5 + 9*C*a^3*b^6 - 3*B *a^2*b^7 - 3*C*a*b^8 + B*b^9)*cos(d*x + c))*log(sin(d*x + c) + 1) - 2*((3* C*a^9 - B*a^8*b - 9*C*a^7*b^2 + 3*B*a^6*b^3 + 9*C*a^5*b^4 - 3*B*a^4*b^5 - 3*C*a^3*b^6 + B*a^2*b^7)*cos(d*x + c)^3 + 2*(3*C*a^8*b - B*a^7*b^2 - 9*C*a ^6*b^3 + 3*B*a^5*b^4 + 9*C*a^4*b^5 - 3*B*a^3*b^6 - 3*C*a^2*b^7 + B*a*b^8)* cos(d*x + c)^2 + (3*C*a^7*b^2 - B*a^6*b^3 - 9*C*a^5*b^4 + 3*B*a^4*b^5 + 9* C*a^3*b^6 - 3*B*a^2*b^7 - 3*C*a*b^8 + B*b^9)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*C*a^6*b^3 - 6*C*a^4*b^5 + 6*C*a^2*b^7 - 2*C*b^9 + (6*C*a^8* b - 2*B*a^7*b^2 - 17*C*a^6*b^3 + 7*B*a^5*b^4 + 13*C*a^4*b^5 - 5*B*a^3*b^6 - 2*C*a^2*b^7)*cos(d*x + c)^2 + (9*C*a^7*b^2 - 3*B*a^6*b^3 - 25*C*a^5*b...
\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \] Input:
integrate(sec(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3 ,x)
Output:
Integral((B + C*sec(c + d*x))*sec(c + d*x)**4/(a + b*sec(c + d*x))**3, x)
Exception generated. \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (277) = 554\).
Time = 0.40 (sec) , antiderivative size = 581, normalized size of antiderivative = 2.01 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")
Output:
((6*C*a^6 - 2*B*a^5*b - 15*C*a^4*b^2 + 5*B*a^3*b^3 + 12*C*a^2*b^4 - 6*B*a* b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/ 2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4*b^4 - 2* a^2*b^6 + b^8)*sqrt(-a^2 + b^2)) - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 2*B*a ^5*b*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^4*b ^2*tan(1/2*d*x + 1/2*c)^3 - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^3*b ^3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^2*b ^4*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^6*tan(1/2*d*x + 1/2*c) + 2*B*a^5*b*tan(1 /2*d*x + 1/2*c) - 5*C*a^5*b*tan(1/2*d*x + 1/2*c) + 3*B*a^4*b^2*tan(1/2*d*x + 1/2*c) + 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c) - 5*B*a^3*b^3*tan(1/2*d*x + 1 /2*c) + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c) - 6*B*a^2*b^4*tan(1/2*d*x + 1/2*c ))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2) - (3*C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/ b^4 + (3*C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*C*tan(1/2*d *x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b^3))/d
Time = 24.91 (sec) , antiderivative size = 9286, normalized size of antiderivative = 32.13 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:
int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x ))^3),x)
Output:
((tan(c/2 + (d*x)/2)^5*(6*C*a^5 - 2*C*b^5 + 6*B*a^2*b^3 + B*a^3*b^2 + 4*C* a^2*b^3 - 12*C*a^3*b^2 - 2*B*a^4*b + 2*C*a*b^4 - 3*C*a^4*b))/((a*b^3 - b^4 )*(a + b)^2) + (tan(c/2 + (d*x)/2)*(6*C*a^5 + 2*C*b^5 + 6*B*a^2*b^3 - B*a^ 3*b^2 - 4*C*a^2*b^3 - 12*C*a^3*b^2 - 2*B*a^4*b + 2*C*a*b^4 + 3*C*a^4*b))/( (a + b)*(b^5 - 2*a*b^4 + a^2*b^3)) - (2*tan(c/2 + (d*x)/2)^3*(6*C*a^6 - 2* C*b^6 + 5*B*a^3*b^3 + 6*C*a^2*b^4 - 13*C*a^4*b^2 - 2*B*a^5*b))/(b*(a*b^2 - b^3)*(a + b)^2*(a - b)))/(d*(2*a*b - tan(c/2 + (d*x)/2)^2*(2*a*b + 3*a^2 - b^2) - tan(c/2 + (d*x)/2)^6*(a^2 - 2*a*b + b^2) + a^2 + b^2 - tan(c/2 + (d*x)/2)^4*(2*a*b - 3*a^2 + b^2))) + (atan((((B*b - 3*C*a)*((8*tan(c/2 + ( d*x)/2)*(4*B^2*b^12 + 72*C^2*a^12 - 8*B^2*a*b^11 - 72*C^2*a^11*b + 24*B^2* a^2*b^10 + 32*B^2*a^3*b^9 - 52*B^2*a^4*b^8 - 48*B^2*a^5*b^7 + 57*B^2*a^6*b ^6 + 32*B^2*a^7*b^5 - 32*B^2*a^8*b^4 - 8*B^2*a^9*b^3 + 8*B^2*a^10*b^2 + 36 *C^2*a^2*b^10 - 72*C^2*a^3*b^9 + 36*C^2*a^4*b^8 + 288*C^2*a^5*b^7 - 288*C^ 2*a^6*b^6 - 432*C^2*a^7*b^5 + 441*C^2*a^8*b^4 + 288*C^2*a^9*b^3 - 288*C^2* a^10*b^2 - 24*B*C*a*b^11 - 48*B*C*a^11*b + 48*B*C*a^2*b^10 - 72*B*C*a^3*b^ 9 - 192*B*C*a^4*b^8 + 252*B*C*a^5*b^7 + 288*B*C*a^6*b^6 - 318*B*C*a^7*b^5 - 192*B*C*a^8*b^4 + 192*B*C*a^9*b^3 + 48*B*C*a^10*b^2))/(a*b^12 + b^13 - 3 *a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) + (((8 *(4*B*b^18 - 8*B*a^2*b^16 + 34*B*a^3*b^15 + 6*B*a^4*b^14 - 36*B*a^5*b^13 - 4*B*a^6*b^12 + 18*B*a^7*b^11 + 2*B*a^8*b^10 - 4*B*a^9*b^9 + 24*C*a^2*b...
Time = 0.17 (sec) , antiderivative size = 3960, normalized size of antiderivative = 13.70 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)
Output:
(12*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt( - a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**2*a**8*c - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2) )*cos(c + d*x)*sin(c + d*x)**2*a**7*b**2 - 30*sqrt( - a**2 + b**2)*atan((t an((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x) *sin(c + d*x)**2*a**6*b**2*c + 10*sqrt( - a**2 + b**2)*atan((tan((c + d*x) /2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*sin(c + d*x )**2*a**5*b**4 + 24*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**2*a**4*b**4 *c - 12*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b )/sqrt( - a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**2*a**3*b**6 - 12*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**8*c + 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x) /2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**7*b**2 + 18*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt( - a**2 + b**2))*cos(c + d*x)*a**6*b**2*c - 6*sqrt( - a**2 + b**2)*atan ((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d *x)*a**5*b**4 + 6*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**4*b**4*c + 2*sqrt( - a** 2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 +...