Optimal. Leaf size=441 \[ -\frac{2 \sqrt{2 \pi } e^2 \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{6 \sqrt{6 \pi } e^2 \sin \left (\frac{3 a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{5 b^{7/2} d}+\frac{2 \sqrt{2 \pi } e^2 \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{6 \sqrt{6 \pi } e^2 \cos \left (\frac{3 a}{b}\right ) S\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{5 b^{7/2} d}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{24 e^2 \sqrt{1-(c+d x)^2} (c+d x)^2}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^2 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 e^2 \sqrt{1-(c+d x)^2} (c+d x)^2}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
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Rubi [A] time = 1.19786, antiderivative size = 441, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {4805, 12, 4633, 4719, 4631, 3306, 3305, 3351, 3304, 3352, 4621, 4723} \[ -\frac{2 \sqrt{2 \pi } e^2 \sin \left (\frac{a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{6 \sqrt{6 \pi } e^2 \sin \left (\frac{3 a}{b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{5 b^{7/2} d}+\frac{2 \sqrt{2 \pi } e^2 \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{6 \sqrt{6 \pi } e^2 \cos \left (\frac{3 a}{b}\right ) S\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{5 b^{7/2} d}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{24 e^2 \sqrt{1-(c+d x)^2} (c+d x)^2}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^2 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{2 e^2 \sqrt{1-(c+d x)^2} (c+d x)^2}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
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
Rule 4621
Rule 4723
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^2}{\left (a+b \sin ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^2 x^2}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac{\left (4 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}-\frac{\left (6 e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\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^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{\left (8 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}-\frac{\left (12 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{5 b^2 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^2 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{24 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{\left (16 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x^2} \sqrt{a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{15 b^3 d}-\frac{\left (24 e^2\right ) \operatorname{Subst}\left (\int \left (-\frac{\sin (x)}{4 \sqrt{a+b x}}+\frac{3 \sin (3 x)}{4 \sqrt{a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^2 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{24 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{\left (16 e^2\right ) \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (6 e^2\right ) \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}-\frac{\left (18 e^2\right ) \operatorname{Subst}\left (\int \frac{\sin (3 x)}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^2 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{24 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{\left (16 e^2 \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 )}{15 b^3 d}+\frac{\left (6 e^2 \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 )}{5 b^3 d}-\frac{\left (18 e^2 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}+\frac{\left (16 e^2 \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 )}{15 b^3 d}-\frac{\left (6 e^2 \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 )}{5 b^3 d}+\frac{\left (18 e^2 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{\sqrt{a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{5 b^3 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^2 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{24 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}-\frac{\left (32 e^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 )}{15 b^4 d}+\frac{\left (12 e^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 )}{5 b^4 d}-\frac{\left (36 e^2 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \sin \left (\frac{3 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{5 b^4 d}+\frac{\left (32 e^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 )}{15 b^4 d}-\frac{\left (12 e^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 )}{5 b^4 d}+\frac{\left (36 e^2 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \cos \left (\frac{3 x^2}{b}\right ) \, dx,x,\sqrt{a+b \sin ^{-1}(c+d x)}\right )}{5 b^4 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac{8 e^2 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac{4 e^2 (c+d x)^3}{5 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^2 \sqrt{1-(c+d x)^2}}{15 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{24 e^2 (c+d x)^2 \sqrt{1-(c+d x)^2}}{5 b^3 d \sqrt{a+b \sin ^{-1}(c+d x)}}+\frac{2 e^2 \sqrt{2 \pi } \cos \left (\frac{a}{b}\right ) S\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{6 e^2 \sqrt{6 \pi } \cos \left (\frac{3 a}{b}\right ) S\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right )}{5 b^{7/2} d}-\frac{2 e^2 \sqrt{2 \pi } C\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right ) \sin \left (\frac{a}{b}\right )}{15 b^{7/2} d}+\frac{6 e^2 \sqrt{6 \pi } C\left (\frac{\sqrt{\frac{6}{\pi }} \sqrt{a+b \sin ^{-1}(c+d x)}}{\sqrt{b}}\right ) \sin \left (\frac{3 a}{b}\right )}{5 b^{7/2} d}\\ \end{align*}
Mathematica [C] time = 1.77886, size = 538, normalized size = 1.22 \[ \frac{e^2 \left (e^{-i \sin ^{-1}(c+d x)} \left (-4 e^{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^2 \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+4 a^2+2 a b \left (4 \sin ^{-1}(c+d x)+i\right )+b^2 \left (4 \sin ^{-1}(c+d x)^2+2 i \sin ^{-1}(c+d x)-3\right )\right )+3 e^{-3 i \sin ^{-1}(c+d x)} \left (b^2-2 \left (a+b \sin ^{-1}(c+d x)\right ) \left (6 i \sqrt{3} b e^{\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+6 a+6 b \sin ^{-1}(c+d x)+i b\right )\right )+2 e^{-\frac{i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right ) \left (e^{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (2 a+2 b \sin ^{-1}(c+d x)-i b\right )-2 i b \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )-6 e^{-\frac{3 i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right ) \left (e^{\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (6 a+6 b \sin ^{-1}(c+d x)-i b\right )-6 i \sqrt{3} b \left (-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )-3 b^2 e^{i \sin ^{-1}(c+d x)}+3 b^2 e^{3 i \sin ^{-1}(c+d x)}\right )}{60 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.155, size = 1229, normalized size = 2.8 \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 )}^{2}}{{\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 )}^{2}}{{\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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