Optimal. Leaf size=204 \[ -\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{4 d}+\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{4 d}-\frac{15 b^2 (c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{4 d}+\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d} \]
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Rubi [A] time = 0.415817, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {4803, 4619, 4677, 4723, 3306, 3305, 3351, 3304, 3352} \[ -\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{4 d}+\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{4 d}-\frac{15 b^2 (c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{4 d}+\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d} \]
Antiderivative was successfully verified.
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Rule 4803
Rule 4619
Rule 4677
Rule 4723
Rule 3306
Rule 3305
Rule 3351
Rule 3304
Rule 3352
Rubi steps
\begin{align*} \int \left (a+b \sin ^{-1}(c+d x)\right )^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \sin ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{x \left (a+b \sin ^{-1}(x)\right )^{3/2}}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d}-\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \sqrt{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{4 d}\\ &=-\frac{15 b^2 (c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{4 d}+\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d}+\frac{\left (15 b^3\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sqrt{a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{8 d}\\ &=-\frac{15 b^2 (c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{4 d}+\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d}+\frac{\left (15 b^3\right ) \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}\\ &=-\frac{15 b^2 (c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{4 d}+\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d}+\frac{\left (15 b^3 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}-\frac{\left (15 b^3 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}\\ &=-\frac{15 b^2 (c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{4 d}+\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d}+\frac{\left (15 b^2 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{4 d}-\frac{\left (15 b^2 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{4 d}\\ &=-\frac{15 b^2 (c+d x) \sqrt{a+b \sin ^{-1}(c+d x)}}{4 d}+\frac{5 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}{2 d}+\frac{(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}{d}+\frac{15 b^{5/2} \sqrt{\frac{\pi }{2}} \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{4 d}-\frac{15 b^{5/2} \sqrt{\frac{\pi }{2}} C\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right ) \sin \left (\frac{a}{b}\right )}{4 d}\\ \end{align*}
Mathematica [C] time = 1.57182, size = 432, normalized size = 2.12 \[ \frac{e^{-\frac{i a}{b}} \left (2 b \left (2 a^2 \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{3}{2},-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 a^2 e^{\frac{2 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{3}{2},\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{\frac{i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right ) \left (2 \sin ^{-1}(c+d x) \left (4 a (c+d x)+5 b \sqrt{-c^2-2 c d x-d^2 x^2+1}\right )+10 a \sqrt{-c^2-2 c d x-d^2 x^2+1}-15 b (c+d x)+4 b (c+d x) \sin ^{-1}(c+d x)^2\right )\right )+\frac{i \sqrt{\frac{\pi }{2}} \left (4 a^2+15 b^2\right ) \left (-1+e^{\frac{2 i a}{b}}\right ) \sqrt{a+b \sin ^{-1}(c+d x)} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\frac{1}{b}} \sqrt{a+b \sin ^{-1}(c+d x)}\right )}{\sqrt{\frac{1}{b}}}+\frac{\sqrt{\frac{\pi }{2}} \left (4 a^2+15 b^2\right ) \left (1+e^{\frac{2 i a}{b}}\right ) \sqrt{a+b \sin ^{-1}(c+d x)} S\left (\sqrt{\frac{1}{b}} \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}\right )}{\sqrt{\frac{1}{b}}}\right )}{8 d \sqrt{a+b \sin ^{-1}(c+d x)}} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0., size = 433, normalized size = 2.1 \begin{align*}{\frac{1}{8\,d} \left ( 15\,\sqrt{{b}^{-1}}\sqrt{\pi }\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }\cos \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ){b}^{3}-15\,\sqrt{{b}^{-1}}\sqrt{\pi }\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }\sin \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{{b}^{-1}}\sqrt{\pi }b}} \right ){b}^{3}+8\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{3}\sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ){b}^{3}+24\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}\sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ) a{b}^{2}+20\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}\cos \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ){b}^{3}+24\,\arcsin \left ( dx+c \right ) \sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ){a}^{2}b-30\,\arcsin \left ( dx+c \right ) \sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ){b}^{3}+40\,\arcsin \left ( dx+c \right ) \cos \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ) a{b}^{2}+8\,\sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ){a}^{3}-30\,\sin \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ) a{b}^{2}+20\,\cos \left ({\frac{a+b\arcsin \left ( dx+c \right ) }{b}}-{\frac{a}{b}} \right ){a}^{2}b \right ){\frac{1}{\sqrt{a+b\arcsin \left ( dx+c \right ) }}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 2.47833, size = 1766, normalized size = 8.66 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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