Integrand size = 45, antiderivative size = 273 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 (A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d} \] Output:
-2/15*(a^2-b^2)*(2*A*b-5*B*a)*((b+a*cos(d*x+c))/(a+b))^(1/2)*InverseJacobi AM(1/2*d*x+1/2*c,2^(1/2)*(a/(a+b))^(1/2))/a^2/d/cos(d*x+c)^(1/2)/(a+b*sec( d*x+c))^(1/2)-2/15*(2*A*b^2-5*B*a*b-3*a^2*(3*A+5*C))*cos(d*x+c)^(1/2)*Elli pticE(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*cos(d*x+c))/(a+b))^(1/2)+2/15*(A*b+5*B*a)*cos(d*x+c)^(1/2)*(a+b *sec(d*x+c))^(1/2)*sin(d*x+c)/a/d+2/5*A*cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^ (1/2)*sin(d*x+c)/d
Result contains complex when optimal does not.
Time = 22.00 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.35 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \left (a (A b+5 a B+3 a A \cos (c+d x)) \sin (c+d x)-\frac {\left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-i (a+b) \left (-2 A b^2+5 a b B+3 a^2 (3 A+5 C)\right ) E\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+i a (a+b) (9 a A-2 A b+5 a (B+3 C)) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-\left (-2 A b^2+5 a b B+3 a^2 (3 A+5 C)\right ) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{(b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}\right )}{15 a^2 d} \] Input:
Integrate[Cos[c + d*x]^(5/2)*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(2*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*(a*(A*b + 5*a*B + 3*a*A*Cos [c + d*x])*Sin[c + d*x] - ((Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*((-I)*( a + b)*(-2*A*b^2 + 5*a*b*B + 3*a^2*(3*A + 5*C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x ])*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(9*a*A - 2*A*b + 5*a*(B + 3* C))*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x) /2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - (-2*A*b^2 + 5*a*b*B + 3*a^2*(3*A + 5*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^( 3/2)*Tan[(c + d*x)/2]))/((b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2))))/(15*a^ 2*d)
Time = 2.42 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.11, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 4753, 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.
\(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^{5/2} \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\) |
\(\Big \downarrow \) 4753 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \sec (c+d x)} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{\sec ^{\frac {5}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4582 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{5} \int \frac {b (2 A+5 C) \sec ^2(c+d x)+(3 a A+5 b B+5 a C) \sec (c+d x)+A b+5 a B}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {b (2 A+5 C) \sec ^2(c+d x)+(3 a A+5 b B+5 a C) \sec (c+d x)+A b+5 a B}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {b (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+5 b B+5 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+A b+5 a B}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \sec (c+d x) a+2 A b^2}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \sec (c+d x) a+2 A b^2}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4523 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{a}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4343 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 4345 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )\) |
Input:
Int[Cos[c + d*x]^(5/2)*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[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)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*S qrt[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)
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]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(2029\) vs. \(2(254)=508\).
Time = 11.30 (sec) , antiderivative size = 2030, normalized size of antiderivative = 7.44
Input:
int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x,method=_RETURNVERBOSE)
Output:
2/15/d*((9*cos(d*x+c)^2+18*cos(d*x+c)+9)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+ b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*EllipticE(((a-b)/(a+b))^(1/2 )*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-9*cos(d*x+c)^2-18*cos(d* x+c)-9)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1 ))^(1/2)*a^2*b*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+ b)/(a-b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*A*(1/(cos(d*x+c)+1))^(1/ 2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE(((a-b)/ (a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(2*cos(d*x+c)^2 +4*cos(d*x+c)+2)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos (d*x+c)+1))^(1/2)*b^3*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c) ),(-(a+b)/(a-b))^(1/2))+(5*cos(d*x+c)^2+10*cos(d*x+c)+5)*B*(1/(cos(d*x+c)+ 1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticE( ((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-5*cos( d*x+c)^2-10*cos(d*x+c)-5)*B*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x +c))/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c) -cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(15*cos(d*x+c)^2+30*cos(d*x+c)+15)*C*(1 /(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3 *EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2 ))+(-15*cos(d*x+c)^2-30*cos(d*x+c)-15)*C*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b) *(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticE(((a-b)/(a+b))^(...
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 531, normalized size of antiderivative = 1.95 \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="fricas")
Output:
2/45*(3*(3*A*a^3*cos(d*x + c) + 5*B*a^3 + A*a^2*b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - sqrt(1/2)*(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)*weierstrassPInver se(-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(1/2)*(-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) - 3*sqrt(1/2)*(-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)) - 3*sqrt(1/2)*(3*I*(3*A + 5*C)*a^3 + 5*I*B*a^2*b - 2*I*A*a*b^2)*sqrt(a)*wei erstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierst rassPInverse(-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)))/(a^3*d)
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)*(a+b*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)+C*sec( d*x+c)**2),x)
Output:
Timed out
\[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a) *cos(d*x + c)^(5/2), x)
\[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a) *cos(d*x + c)^(5/2), x)
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:
int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/ cos(c + d*x)^2),x)
Output:
int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/ cos(c + d*x)^2), x)
\[ \int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a \] Input:
int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x)
Output:
int(sqrt(sec(c + d*x)*b + a)*sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d* x)**2,x)*c + int(sqrt(sec(c + d*x)*b + a)*sqrt(cos(c + d*x))*cos(c + d*x)* *2*sec(c + d*x),x)*b + int(sqrt(sec(c + d*x)*b + a)*sqrt(cos(c + d*x))*cos (c + d*x)**2,x)*a