Integrand size = 27, antiderivative size = 481 \[ \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 a (a-b) \sqrt {a+b} \left (49 A b^2+3 a^2 C+29 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{21 b^2 d}-\frac {2 \sqrt {a+b} \left (3 a^3 C-9 a^2 b (7 A+3 C)-b^3 (7 A+5 C)+a b^2 (49 A+29 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{21 b d}-\frac {2 a^2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 \left (3 a^2 C+b^2 (7 A+5 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{21 d}+\frac {2 a C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \]
-2/21*a*(a-b)*(49*A*b^2+3*C*a^2+29*C*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d* x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c)) /(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d-2/21*(3*a^3*C-9*a^2*b* (7*A+3*C)-b^3*(7*A+5*C)+a*b^2*(49*A+29*C))*cot(d*x+c)*EllipticF((a+b*sec(d *x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c) )/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d-2*a^2*A*cot(d*x+c)*Elli pticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+ b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/d+ 2/7*a*C*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/7*C*(a+b*sec(d*x+c))^(5/2)*t an(d*x+c)/d+2/21*(3*C*a^2+b^2*(7*A+5*C))*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c) /d
Leaf count is larger than twice the leaf count of optimal. \(4075\) vs. \(2(481)=962\).
Time = 28.46 (sec) , antiderivative size = 4075, normalized size of antiderivative = 8.47 \[ \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]
(Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2)*((4*a*(4 9*A*b^2 + 3*a^2*C + 29*b^2*C)*Sin[c + d*x])/(21*b) + (4*Sec[c + d*x]*(7*A* b^2*Sin[c + d*x] + 9*a^2*C*Sin[c + d*x] + 5*b^2*C*Sin[c + d*x]))/21 + (12* a*b*C*Sec[c + d*x]*Tan[c + d*x])/7 + (4*b^2*C*Sec[c + d*x]^2*Tan[c + d*x]) /7))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + A*Cos[2*c + 2*d*x])) + (4*((2*a^ 3*A)/(Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (14*a*A*b^2)/(3*Sqrt[ b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^3*C)/(7*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (58*a*b^2*C)/(21*Sqrt[b + a*Cos[c + d*x]]*Sqrt [Sec[c + d*x]]) + (4*a^2*A*b*Sqrt[Sec[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x ]]) + (2*A*b^3*Sqrt[Sec[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]) - (2*a^4*C *Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (4*a^2*b*C*Sqrt[Sec[ c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) + (10*b^3*C*Sqrt[Sec[c + d*x]])/( 21*Sqrt[b + a*Cos[c + d*x]]) - (14*a^2*A*b*Cos[2*(c + d*x)]*Sqrt[Sec[c + d *x]])/(3*Sqrt[b + a*Cos[c + d*x]]) - (2*a^4*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (58*a^2*b*C*Cos[2*(c + d*x)]*Sqr t[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Cos[(c + d*x)/2]^2*Se c[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2)*(-2*a*(a + b )*(49*A*b^2 + 3*a^2*C + 29*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sq rt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan [(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(3*a^3*(-7*A + C) + 9*a^2*b*(7*A...
Time = 2.02 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.630, Rules used = {3042, 4545, 27, 3042, 4544, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
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.
\(\displaystyle \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4545 |
\(\displaystyle \frac {2}{7} \int \frac {1}{2} (a+b \sec (c+d x))^{3/2} \left (5 a C \sec ^2(c+d x)+b (7 A+5 C) \sec (c+d x)+7 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \int (a+b \sec (c+d x))^{3/2} \left (5 a C \sec ^2(c+d x)+b (7 A+5 C) \sec (c+d x)+7 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (7 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )+7 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 4544 |
\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {5}{2} \sqrt {a+b \sec (c+d x)} \left (7 A a^2+2 b (7 A+4 C) \sec (c+d x) a+\left (3 C a^2+b^2 (7 A+5 C)\right ) \sec ^2(c+d x)\right )dx+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\int \sqrt {a+b \sec (c+d x)} \left (7 A a^2+2 b (7 A+4 C) \sec (c+d x) a+\left (3 C a^2+b^2 (7 A+5 C)\right ) \sec ^2(c+d x)\right )dx+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\int \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (7 A a^2+2 b (7 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (3 C a^2+b^2 (7 A+5 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 4544 |
\(\displaystyle \frac {1}{7} \left (\frac {2}{3} \int \frac {21 A a^3+\left (3 C a^2+49 A b^2+29 b^2 C\right ) \sec ^2(c+d x) a+b \left (9 (7 A+3 C) a^2+b^2 (7 A+5 C)\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \frac {21 A a^3+\left (3 C a^2+49 A b^2+29 b^2 C\right ) \sec ^2(c+d x) a+b \left (9 (7 A+3 C) a^2+b^2 (7 A+5 C)\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \frac {21 A a^3+\left (3 C a^2+49 A b^2+29 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 a+b \left (9 (7 A+3 C) a^2+b^2 (7 A+5 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 4546 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {21 A a^3+\left (b \left (9 (7 A+3 C) a^2+b^2 (7 A+5 C)\right )-a \left (3 C a^2+49 A b^2+29 b^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {21 A a^3+\left (b \left (9 (7 A+3 C) a^2+b^2 (7 A+5 C)\right )-a \left (3 C a^2+49 A b^2+29 b^2 C\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 4409 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (21 a^3 A \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (3 a^3 C-9 a^2 b (7 A+3 C)+a b^2 (49 A+29 C)-b^3 (7 A+5 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (21 a^3 A \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (3 a^3 C-9 a^2 b (7 A+3 C)+a b^2 (49 A+29 C)-b^3 (7 A+5 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 4271 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (3 a^3 C-9 a^2 b (7 A+3 C)+a b^2 (49 A+29 C)-b^3 (7 A+5 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {42 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (a \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {42 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (3 a^3 C-9 a^2 b (7 A+3 C)+a b^2 (49 A+29 C)-b^3 (7 A+5 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {1}{7} \left (\frac {2 \left (3 a^2 C+7 A b^2+5 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (-\frac {2 a (a-b) \sqrt {a+b} \left (3 a^2 C+49 A b^2+29 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {42 a^2 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 \sqrt {a+b} \left (3 a^3 C-9 a^2 b (7 A+3 C)+a b^2 (49 A+29 C)-b^3 (7 A+5 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}\) |
(2*C*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + (((-2*a*(a - b)*Sqrt [a + b]*(49*A*b^2 + 3*a^2*C + 29*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt [a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d *x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*Sqrt[ a + b]*(3*a^3*C - 9*a^2*b*(7*A + 3*C) - b^3*(7*A + 5*C) + a*b^2*(49*A + 29 *C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (42*a^2*A*Sqrt[a + b]*Cot[c + d*x]*EllipticPi [(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)] *Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b ))])/d)/3 + (2*(7*A*b^2 + 3*a^2*C + 5*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[ c + d*x])/(3*d) + (2*a*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/d)/7
3.8.28.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[( a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m )*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^ m/(f*(m + 1))), x] + Simp[1/(m + 1) Int[(a + b*Csc[e + f*x])^(m - 1)*Simp [a*A*(m + 1) + (A*b*(m + 1) + b*C*m)*Csc[e + f*x] + a*C*m*Csc[e + f*x]^2, x ], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(4266\) vs. \(2(438)=876\).
Time = 34.60 (sec) , antiderivative size = 4267, normalized size of antiderivative = 8.87
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4267\) |
default | \(\text {Expression too large to display}\) | \(4311\) |
2/3*A/d*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(sin(d*x+c) *cos(d*x+c)*a*b^2+14*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))* (1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1) )^(1/2)*a^2*b*cos(d*x+c)+sin(d*x+c)*b^3+8*a*b^2*sin(d*x+c)+3*(1/(a+b)*(b+a *cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellip ticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)^2+6*(1/(a+b )*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2) *EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*cos(d*x+c)+b^3*t an(d*x+c)-6*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(1/(a +b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/ 2)*a^3*cos(d*x+c)^2-12*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^( 1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x +c)+1))^(1/2)*a^3*cos(d*x+c)+7*a^2*b*cos(d*x+c)*sin(d*x+c)-(1/(a+b)*(b+a*c os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Ellipti cF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*b^3+14*EllipticE(cot(d*x+c)- csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^ (1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*cos(d*x+c)-18*EllipticF(cot( d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+ c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)-14*Ellipti cF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))...
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
integral((C*b^2*sec(d*x + c)^4 + 2*C*a*b*sec(d*x + c)^3 + 2*A*a*b*sec(d*x + c) + A*a^2 + (C*a^2 + A*b^2)*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x )
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}\, dx \]
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
\[ \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
Timed out. \[ \int (a+b \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]