Integrand size = 45, antiderivative size = 273 \[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{15 a^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+5 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}} \]
-2/15*(a^2-b^2)*(2*A*b-5*B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2 *c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c) )/(a+b))^(1/2)*sec(d*x+c)^(1/2)/a^2/d/(a+b*sec(d*x+c))^(1/2)+2/5*A*sin(d*x +c)*(a+b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(3/2)+2/15*(A*b+5*B*a)*sin(d*x+c)* (a+b*sec(d*x+c))^(1/2)/a/d/sec(d*x+c)^(1/2)-2/15*(2*A*b^2-5*B*a*b-3*a^2*(3 *A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2 *d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/a^2/d/((b+a*co s(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 7.34 (sec) , antiderivative size = 3426, normalized size of antiderivative = 12.55 \[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Result too large to show} \]
Integrate[(Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2 ))/Sec[c + d*x]^(5/2),x]
(Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(9* a^2*A - 2*A*b^2 + 5*a*b*B + 15*a^2*C)*Cot[c])/(15*a^2*d) + (4*(A*b + 5*a*B )*Cos[d*x]*Sin[c])/(15*a*d) + (2*A*Cos[2*d*x]*Sin[2*c])/(5*d) + (4*(A*b + 5*a*B)*Cos[c]*Sin[d*x])/(15*a*d) + (2*A*Cos[2*c]*Sin[2*d*x])/(5*d)))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(5/2)) - (28*A* b*AppellF1[1/2, 1/2, 1/2, 3/2, (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Si n[d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))), (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[ Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(-1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))) ]*Csc[c]*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)* Sec[d*x - ArcTan[Cot[c]]]*Sqrt[(a*Sqrt[1 + Cot[c]^2] - a*Sqrt[1 + Cot[c]^2 ]*Sin[d*x - ArcTan[Cot[c]]])/(a*Sqrt[1 + Cot[c]^2] - b*Csc[c])]*Sqrt[(a*Sq rt[1 + Cot[c]^2] + a*Sqrt[1 + Cot[c]^2]*Sin[d*x - ArcTan[Cot[c]]])/(a*Sqrt [1 + Cot[c]^2] + b*Csc[c])]*Sqrt[b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]])/(15*a*d*Sqrt[b + a*Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*Sec[c + d*x]^(5/2)) - (4*B* AppellF1[1/2, 1/2, 1/2, 3/2, (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[ d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))), (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Co t[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(-1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))...
Time = 2.10 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.03, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.422, Rules used = {3042, 4582, 27, 3042, 4592, 27, 3042, 4523, 3042, 4343, 3042, 3134, 3042, 3132, 4345, 3042, 3142, 3042, 3140}
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.
\(\displaystyle \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4582 |
\(\displaystyle \frac {2}{5} \int \frac {b (2 A+5 C) \sec ^2(c+d x)+(3 a A+5 b B+5 a C) \sec (c+d x)+A b+5 a B}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {b (2 A+5 C) \sec ^2(c+d x)+(3 a A+5 b B+5 a C) \sec (c+d x)+A b+5 a B}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {b (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+5 b B+5 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+A b+5 a B}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \sec (c+d x) a+2 A b^2}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \sec (c+d x) a+2 A b^2}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4523 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{a}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4343 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4345 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
(2*A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (-1 /3*((2*(a^2 - b^2)*(2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Elli pticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(a*d*Sqrt[a + b*Sec[ c + d*x]]) + (2*(2*A*b^2 - 5*a*b*B - 3*a^2*(3*A + 5*C))*EllipticE[(c + d*x )/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(a*d*Sqrt[(b + a*Cos[c + d*x ])/(a + b)]*Sqrt[Sec[c + d*x]]))/a + (2*(A*b + 5*a*B)*Sqrt[a + b*Sec[c + d *x]]*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]))/5
3.11.35.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[Sqrt[a + b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*S qrt[b + a*Sin[e + f*x]]) Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[{a , b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/S qrt[a + b*Csc[e + f*x]]) Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[ {a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d _.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Simp[A/a I nt[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Simp[(A*b - a*B) /(a*d) Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ [{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(4449\) vs. \(2(303)=606\).
Time = 11.11 (sec) , antiderivative size = 4450, normalized size of antiderivative = 16.30
method | result | size |
parts | \(\text {Expression too large to display}\) | \(4450\) |
default | \(\text {Expression too large to display}\) | \(4802\) |
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2 ),x,method=_RETURNVERBOSE)
-2/15*A/d/a^2/((a-b)/(a+b))^(1/2)*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/ sec(d*x+c)^(5/2)/(1+cos(d*x+c))*(-7*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c )))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x +c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b-2*(1/(a+b)*(b+a*cos(d*x+c))/(1 +cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2) *(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2+9*(1/(a+b)*(b+a*cos( d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a +b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b+2*(1/(a+b) *(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE (((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2+ 9*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2) *EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/ 2))*a^3*sec(d*x+c)^2-9*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/ (1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c )),(-(a+b)/(a-b))^(1/2))*a^3*sec(d*x+c)^2-2*(1/(a+b)*(b+a*cos(d*x+c))/(1+c os(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*( -cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b^3*sec(d*x+c)^2+18*(1/(a+b) *(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF (((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*se c(d*x+c)-18*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 539, normalized size of antiderivative = 1.97 \[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left (-15 i \, B a^{3} - 3 i \, {\left (A + 5 \, C\right )} a^{2} b + 10 i \, B a b^{2} - 4 i \, A b^{3}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) + \sqrt {2} {\left (15 i \, B a^{3} + 3 i \, {\left (A + 5 \, C\right )} a^{2} b - 10 i \, B a b^{2} + 4 i \, A b^{3}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) - 3 \, \sqrt {2} {\left (-3 i \, {\left (3 \, A + 5 \, C\right )} a^{3} - 5 i \, B a^{2} b + 2 i \, A a b^{2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, {\left (3 \, A + 5 \, C\right )} a^{3} + 5 i \, B a^{2} b - 2 i \, A a b^{2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) + \frac {6 \, {\left (3 \, A a^{3} \cos \left (d x + c\right )^{2} + {\left (5 \, B a^{3} + A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{45 \, a^{3} d} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2)/sec(d*x+c )^(5/2),x, algorithm="fricas")
1/45*(sqrt(2)*(-15*I*B*a^3 - 3*I*(A + 5*C)*a^2*b + 10*I*B*a*b^2 - 4*I*A*b^ 3)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8 *b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) + sqrt(2)* (15*I*B*a^3 + 3*I*(A + 5*C)*a^2*b - 10*I*B*a*b^2 + 4*I*A*b^3)*sqrt(a)*weie rstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3* (3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 3*sqrt(2)*(-3*I*(3*A + 5*C)*a^3 - 5*I*B*a^2*b + 2*I*A*a*b^2)*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*si n(d*x + c) + 2*b)/a)) - 3*sqrt(2)*(3*I*(3*A + 5*C)*a^3 + 5*I*B*a^2*b - 2*I *A*a*b^2)*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a)) + 6*(3 *A*a^3*cos(d*x + c)^2 + (5*B*a^3 + A*a^2*b)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d)
\[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a + b \sec {\left (c + d x \right )}} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
Integral(sqrt(a + b*sec(c + d*x))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2) /sec(c + d*x)**(5/2), x)
\[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2)/sec(d*x+c )^(5/2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a) /sec(d*x + c)^(5/2), x)
\[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2)/sec(d*x+c )^(5/2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a) /sec(d*x + c)^(5/2), x)
Timed out. \[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
int(((a + b/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 1/cos(c + d*x))^(5/2),x)