Optimal. Leaf size=442 \[ \frac{32 \sqrt{2 \pi } e^3 \cos \left (\frac{4 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{16 \sqrt{\pi } e^3 \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{15 b^{7/2} d}-\frac{16 \sqrt{\pi } e^3 \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{15 b^{7/2} d}+\frac{32 \sqrt{2 \pi } e^3 \sin \left (\frac{4 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{128 e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 \sqrt{1-(c+d x)^2} (c+d x)}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
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Rubi [A] time = 1.13726, antiderivative size = 442, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4805, 12, 4633, 4719, 4631, 3306, 3305, 3351, 3304, 3352} \[ \frac{32 \sqrt{2 \pi } e^3 \cos \left (\frac{4 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{16 \sqrt{\pi } e^3 \cos \left (\frac{2 a}{b}\right ) \text{FresnelC}\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{\pi } \sqrt{b}}\right )}{15 b^{7/2} d}-\frac{16 \sqrt{\pi } e^3 \sin \left (\frac{2 a}{b}\right ) S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{15 b^{7/2} d}+\frac{32 \sqrt{2 \pi } e^3 \sin \left (\frac{4 a}{b}\right ) S\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{128 e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 \sqrt{1-(c+d x)^2} (c+d x)}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 12
Rule 4633
Rule 4719
Rule 4631
Rule 3306
Rule 3305
Rule 3351
Rule 3304
Rule 3352
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{\left (6 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{5 b^2 d}-\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{128 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{\left (16 e^3\right ) \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}-\frac{\left (128 e^3\right ) \operatorname{Subst}\left (\int \left (\frac{\cos (2 x)}{2 \sqrt{a+b x}}-\frac{\cos (4 x)}{2 \sqrt{a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{128 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{\cos (4 x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (16 e^3 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}+\frac{\left (16 e^3 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{128 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{\left (32 e^3 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{5 b^4 d}-\frac{\left (64 e^3 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (64 e^3 \cos \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{4 a}{b}+4 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (32 e^3 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{5 b^4 d}-\frac{\left (64 e^3 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (64 e^3 \sin \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{4 a}{b}+4 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{128 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{16 e^3 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) C\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{5 b^{7/2} d}+\frac{16 e^3 \sqrt{\pi } S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right ) \sin \left (\frac{2 a}{b}\right )}{5 b^{7/2} d}-\frac{\left (128 e^3 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}+\frac{\left (128 e^3 \cos \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{4 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}-\frac{\left (128 e^3 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{2 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}+\frac{\left (128 e^3 \sin \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{4 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{128 e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{32 e^3 \sqrt{2 \pi } \cos \left (\frac{4 a}{b}\right ) C\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{16 e^3 \sqrt{\pi } \cos \left (\frac{2 a}{b}\right ) C\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right )}{15 b^{7/2} d}-\frac{16 e^3 \sqrt{\pi } S\left (\frac{2 \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b} \sqrt{\pi }}\right ) \sin \left (\frac{2 a}{b}\right )}{15 b^{7/2} d}+\frac{32 e^3 \sqrt{2 \pi } S\left (\frac{2 \sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right ) \sin \left (\frac{4 a}{b}\right )}{15 b^{7/2} d}\\ \end{align*}
Mathematica [C] time = 2.286, size = 445, normalized size = 1.01 \[ \frac{e^3 \left (-4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (4 \sqrt{2} b e^{-\frac{2 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{-2 i \sin ^{-1}(c+d x)} \left (4 \sqrt{2} b e^{\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},\frac{2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-4 i a-4 i b \sin ^{-1}(c+d x)+b\right )+e^{2 i \sin ^{-1}(c+d x)} \left (4 i a+4 i b \sin ^{-1}(c+d x)+b\right )\right )+4 \left (a+b \sin ^{-1}(c+d x)\right ) \left (16 b e^{-\frac{4 i a}{b}} \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+16 b e^{\frac{4 i a}{b}} \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},\frac{4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{-4 i \sin ^{-1}(c+d x)} \left (-8 i a-8 i b \sin ^{-1}(c+d x)+b\right )+e^{4 i \sin ^{-1}(c+d x)} \left (8 i a+8 i b \sin ^{-1}(c+d x)+b\right )\right )-6 b^2 \sin \left (2 \sin ^{-1}(c+d x)\right )+3 b^2 \sin \left (4 \sin ^{-1}(c+d x)\right )\right )}{60 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.148, size = 1181, normalized size = 2.7 \begin{align*} \text{result too large to display} \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 \frac{{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac{7}{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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{3}}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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