Integrand size = 16, antiderivative size = 249 \[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=-\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}-\frac {30 b^2 \sqrt {a+b \arccos \left (-1+d x^2\right )} \cos ^2\left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )}{d x}+\frac {30 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} d x}+\frac {30 \sqrt {\pi } \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{\left (\frac {1}{b}\right )^{5/2} d x} \]
[Out]
Time = 0.04 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {4899, 4897} \[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=-\frac {30 b^2 \cos ^2\left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \sqrt {a+b \arccos \left (d x^2-1\right )}}{d x}-\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a+b \arccos \left (d x^2-1\right )\right )^{3/2}}{d x}+\frac {30 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} d x}+\frac {30 \sqrt {\pi } \sin \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (d x^2-1\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (d x^2-1\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} d x}+x \left (a+b \arccos \left (d x^2-1\right )\right )^{5/2} \]
[In]
[Out]
Rule 4897
Rule 4899
Rubi steps \begin{align*} \text {integral}& = -\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}-\left (15 b^2\right ) \int \sqrt {a+b \arccos \left (-1+d x^2\right )} \, dx \\ & = -\frac {5 b \sqrt {2 d x^2-d^2 x^4} \left (a+b \arccos \left (-1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2}-\frac {30 b^2 \sqrt {a+b \arccos \left (-1+d x^2\right )} \cos ^2\left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )}{d x}+\frac {30 \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelC}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right )}{\left (\frac {1}{b}\right )^{5/2} d x}+\frac {30 \sqrt {\pi } \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \operatorname {FresnelS}\left (\frac {\sqrt {\frac {1}{b}} \sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )}{\left (\frac {1}{b}\right )^{5/2} d x} \\ \end{align*}
Time = 1.45 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.00 \[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=\frac {2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right ) \left (15 b^{5/2} \sqrt {\pi } \cos \left (\frac {a}{2 b}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right )+15 b^{5/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {\sqrt {a+b \arccos \left (-1+d x^2\right )}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {a}{2 b}\right )+\sqrt {a+b \arccos \left (-1+d x^2\right )} \left (\left (a^2-15 b^2\right ) \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )+b^2 \arccos \left (-1+d x^2\right )^2 \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )-5 a b \sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )+b \arccos \left (-1+d x^2\right ) \left (2 a \cos \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )-5 b \sin \left (\frac {1}{2} \arccos \left (-1+d x^2\right )\right )\right )\right )\right )}{d x} \]
[In]
[Out]
\[\int {\left (a +b \arccos \left (d \,x^{2}-1\right )\right )}^{\frac {5}{2}}d x\]
[In]
[Out]
Exception generated. \[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=\text {Exception raised: TypeError} \]
[In]
[Out]
\[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=\int \left (a + b \operatorname {acos}{\left (d x^{2} - 1 \right )}\right )^{\frac {5}{2}}\, dx \]
[In]
[Out]
\[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=\int { {\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
[In]
[Out]
\[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=\int { {\left (b \arccos \left (d x^{2} - 1\right ) + a\right )}^{\frac {5}{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \left (a+b \arccos \left (-1+d x^2\right )\right )^{5/2} \, dx=\int {\left (a+b\,\mathrm {acos}\left (d\,x^2-1\right )\right )}^{5/2} \,d x \]
[In]
[Out]