Integrand size = 33, antiderivative size = 345 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}+\frac {b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \] Output:
(4*A*a^2*b-5*A*b^3-2*B*a^3+3*B*a*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2) )/a^3/(a^2-b^2)/d+1/3*(2*A*a^2-5*A*b^2+3*B*a*b)*InverseJacobiAM(1/2*d*x+1/ 2*c,2^(1/2))/a^2/(a^2-b^2)/d+b*(7*A*a^2*b-5*A*b^3-5*B*a^3+3*B*a*b^2)*Ellip ticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a^3/(a-b)/(a+b)^2/d+1/3*(2*A*a ^2-5*A*b^2+3*B*a*b)*sin(d*x+c)/a^2/(a^2-b^2)/d/cos(d*x+c)^(3/2)-(4*A*a^2*b -5*A*b^3-2*B*a^3+3*B*a*b^2)*sin(d*x+c)/a^3/(a^2-b^2)/d/cos(d*x+c)^(1/2)+b* (A*b-B*a)*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))
Time = 7.35 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.24 \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \left (4 a^4 A+44 a^2 A b^2-45 A b^4-30 a^3 b B+27 a b^3 B\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {\left (28 a^3 A b-40 a A b^3-12 a^4 B+24 a^2 b^2 B\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}+\frac {2 \left (12 a^2 A b^2-15 A b^4-6 a^3 b B+9 a b^3 B\right ) \cos (2 (c+d x)) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {1-\cos ^2(c+d x)} \left (-1+2 \cos ^2(c+d x)\right )}}{12 a^3 (a-b) (a+b) d}+\frac {\sqrt {\cos (c+d x)} \left (\frac {2 \sec (c+d x) (-2 A b \sin (c+d x)+a B \sin (c+d x))}{a^3}+\frac {A b^4 \sin (c+d x)-a b^3 B \sin (c+d x)}{a^3 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {2 A \sec (c+d x) \tan (c+d x)}{3 a^2}\right )}{d} \] Input:
Integrate[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2) ,x]
Output:
((2*(4*a^4*A + 44*a^2*A*b^2 - 45*A*b^4 - 30*a^3*b*B + 27*a*b^3*B)*Elliptic Pi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + ((28*a^3*A*b - 40*a*A*b^3 - 1 2*a^4*B + 24*a^2*b^2*B)*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2* b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b + (2*(12*a^2*A*b^2 - 15*A*b^4 - 6 *a^3*b*B + 9*a*b^3*B)*Cos[2*(c + d*x)]*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + ( -2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[1 - Cos[c + d*x]^2]*(-1 + 2*Cos[c + d*x]^2)))/(12*a^3*(a - b)*(a + b)*d) + (Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]*(-2*A*b*Sin[c + d* x] + a*B*Sin[c + d*x]))/a^3 + (A*b^4*Sin[c + d*x] - a*b^3*B*Sin[c + d*x])/ (a^3*(a^2 - b^2)*(a + b*Cos[c + d*x])) + (2*A*Sec[c + d*x]*Tan[c + d*x])/( 3*a^2)))/d
Time = 2.65 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.95, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 3479, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}
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.
\(\displaystyle \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3479 |
\(\displaystyle \frac {\int \frac {2 A a^2+3 b B a-2 (A b-a B) \cos (c+d x) a-5 A b^2+3 b (A b-a B) \cos ^2(c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 A a^2+3 b B a-2 (A b-a B) \cos (c+d x) a-5 A b^2+3 b (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 A a^2+3 b B a-2 (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a-5 A b^2+3 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {2 \int -\frac {-b \left (2 A a^2+3 b B a-5 A b^2\right ) \cos ^2(c+d x)-2 a \left (A a^2-3 b B a+2 A b^2\right ) \cos (c+d x)+3 \left (-2 B a^3+4 A b a^2+3 b^2 B a-5 A b^3\right )}{2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}+\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-b \left (2 A a^2+3 b B a-5 A b^2\right ) \cos ^2(c+d x)-2 a \left (A a^2-3 b B a+2 A b^2\right ) \cos (c+d x)+3 \left (-2 B a^3+4 A b a^2+3 b^2 B a-5 A b^3\right )}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-b \left (2 A a^2+3 b B a-5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a \left (A a^2-3 b B a+2 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (-2 B a^3+4 A b a^2+3 b^2 B a-5 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {2 A a^4-12 b B a^3+16 A b^2 a^2+9 b^3 B a+2 \left (-3 B a^3+7 A b a^2+6 b^2 B a-10 A b^3\right ) \cos (c+d x) a-15 A b^4+3 b \left (-2 B a^3+4 A b a^2+3 b^2 B a-5 A b^3\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}+\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {2 A a^4-12 b B a^3+16 A b^2 a^2+9 b^3 B a+2 \left (-3 B a^3+7 A b a^2+6 b^2 B a-10 A b^3\right ) \cos (c+d x) a-15 A b^4+3 b \left (-2 B a^3+4 A b a^2+3 b^2 B a-5 A b^3\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\int \frac {2 A a^4-12 b B a^3+16 A b^2 a^2+9 b^3 B a+2 \left (-3 B a^3+7 A b a^2+6 b^2 B a-10 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-15 A b^4+3 b \left (-2 B a^3+4 A b a^2+3 b^2 B a-5 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {a \left (2 A a^2+3 b B a-5 A b^2\right ) \cos (c+d x) b^2+\left (2 A a^4-12 b B a^3+16 A b^2 a^2+9 b^3 B a-15 A b^4\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \int \sqrt {\cos (c+d x)}dx+\frac {\int \frac {a \left (2 A a^2+3 b B a-5 A b^2\right ) \cos (c+d x) b^2+\left (2 A a^4-12 b B a^3+16 A b^2 a^2+9 b^3 B a-15 A b^4\right ) b}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {3 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\int \frac {a \left (2 A a^2+3 b B a-5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (2 A a^4-12 b B a^3+16 A b^2 a^2+9 b^3 B a-15 A b^4\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {\int \frac {a \left (2 A a^2+3 b B a-5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2+\left (2 A a^4-12 b B a^3+16 A b^2 a^2+9 b^3 B a-15 A b^4\right ) b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a b \left (2 a^2 A+3 a b B-5 A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 b^2 \left (-5 a^3 B+7 a^2 A b+3 a b^2 B-5 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {a b \left (2 a^2 A+3 a b B-5 A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 b^2 \left (-5 a^3 B+7 a^2 A b+3 a b^2 B-5 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {3 b^2 \left (-5 a^3 B+7 a^2 A b+3 a b^2 B-5 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a b \left (2 a^2 A+3 a b B-5 A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b}+\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}+\frac {\frac {2 \left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) \sin (c+d x)}{a d \sqrt {\cos (c+d x)}}-\frac {\frac {6 \left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {\frac {2 a b \left (2 a^2 A+3 a b B-5 A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 b^2 \left (-5 a^3 B+7 a^2 A b+3 a b^2 B-5 A b^3\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}}{b}}{a}}{3 a}}{2 a \left (a^2-b^2\right )}\) |
Input:
Int[(A + B*Cos[c + d*x])/(Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2),x]
Output:
(b*(A*b - a*B)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*(a + b*Co s[c + d*x])) + ((2*(2*a^2*A - 5*A*b^2 + 3*a*b*B)*Sin[c + d*x])/(3*a*d*Cos[ c + d*x]^(3/2)) - (-(((6*(4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*Ellip ticE[(c + d*x)/2, 2])/d + ((2*a*b*(2*a^2*A - 5*A*b^2 + 3*a*b*B)*EllipticF[ (c + d*x)/2, 2])/d + (6*b^2*(7*a^2*A*b - 5*A*b^3 - 5*a^3*B + 3*a*b^2*B)*El lipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/b)/a) + (6*(4*a^2*A* b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*Sin[c + d*x])/(a*d*Sqrt[Cos[c + d*x]])) /(3*a))/(2*a*(a^2 - b^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0]) ))
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1004\) vs. \(2(340)=680\).
Time = 8.00 (sec) , antiderivative size = 1005, normalized size of antiderivative = 2.91
Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+cos(d*x+c)*b)^2,x,method=_RETURNV ERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a^2*(-1/6* cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(c os(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d* x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(2*A*b-B*a)/a^3/sin(1/2*d*x+1/2*c) ^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x+1/2*c)^2) ^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/ 2)))-4*b^2*(2*A*b-B*a)/a^3/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2 *cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c) ^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*(A*b-B*a)*b/ a^2*(-b^2/a/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2* d*x+1/2*c)^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/(a+b)*(sin(1/2*d* x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c) ^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*b /(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2 )/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))+1/2*b/(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos( 1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^( 1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)...
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorith m="fricas")
Output:
Timed out
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c))**2,x)
Output:
Timed out
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorith m="maxima")
Output:
Timed out
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int { \frac {B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)^2*cos(d*x + c)^(5/2)) , x)
Timed out. \[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \] Input:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))^2),x)
Output:
int((A + B*cos(c + d*x))/(cos(c + d*x)^(5/2)*(a + b*cos(c + d*x))^2), x)
\[ \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b +\cos \left (d x +c \right )^{3} a}d x \] Input:
int((A+B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x)
Output:
int(sqrt(cos(c + d*x))/(cos(c + d*x)**4*b + cos(c + d*x)**3*a),x)