Optimal. Leaf size=183 \[ \frac{2 b^2 (e (c+d x))^{m+3} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac{2 b (e (c+d x))^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2 (m+1) (m+2)}+\frac{(e (c+d x))^{m+1} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e (m+1)} \]
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Rubi [A] time = 0.198848, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4805, 4627, 4711} \[ \frac{2 b^2 (e (c+d x))^{m+3} \, _3F_2\left (1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2};\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2};(c+d x)^2\right )}{d e^3 (m+1) (m+2) (m+3)}-\frac{2 b (e (c+d x))^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2 (m+1) (m+2)}+\frac{(e (c+d x))^{m+1} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e (m+1)} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 4627
Rule 4711
Rubi steps
\begin{align*} \int (c e+d e x)^m \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int (e x)^m \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{(e (c+d x))^{1+m} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e (1+m)}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{(e x)^{1+m} \left (a+b \sin ^{-1}(x)\right )}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{d e (1+m)}\\ &=\frac{(e (c+d x))^{1+m} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e (1+m)}-\frac{2 b (e (c+d x))^{2+m} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};(c+d x)^2\right )}{d e^2 (1+m) (2+m)}+\frac{2 b^2 (e (c+d x))^{3+m} \, _3F_2\left (1,\frac{3}{2}+\frac{m}{2},\frac{3}{2}+\frac{m}{2};2+\frac{m}{2},\frac{5}{2}+\frac{m}{2};(c+d x)^2\right )}{d e^3 (1+m) (2+m) (3+m)}\\ \end{align*}
Mathematica [A] time = 0.135545, size = 151, normalized size = 0.83 \[ \frac{(c+d x) (e (c+d x))^m \left (\frac{2 b^2 (c+d x)^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},(c+d x)^2\right )}{(m+2) (m+3)}-\frac{2 b (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{m+2}+\left (a+b \sin ^{-1}(c+d x)\right )^2\right )}{d (m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.165, size = 0, normalized size = 0. \begin{align*} \int \left ( dex+ce \right ) ^{m} \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}\right )}{\left (d e x + c e\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \left (c + d x\right )\right )^{m} \left (a + b \operatorname{asin}{\left (c + d x \right )}\right )^{2}\, dx \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{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}{\left (d e x + c e\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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