Optimal. Leaf size=183 \[ \frac {(e (c+d x))^{1+m} (a+b \text {ArcSin}(c+d x))^2}{d e (1+m)}-\frac {2 b (e (c+d x))^{2+m} (a+b \text {ArcSin}(c+d x)) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};(c+d x)^2\right )}{d e^2 (1+m) (2+m)}+\frac {2 b^2 (e (c+d x))^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};(c+d x)^2\right )}{d e^3 (1+m) (2+m) (3+m)} \]
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Rubi [A]
time = 0.15, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4889, 4723,
4805} \begin {gather*} \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 ) (a+b \text {ArcSin}(c+d x))}{d e^2 (m+1) (m+2)}+\frac {(e (c+d x))^{m+1} (a+b \text {ArcSin}(c+d x))^2}{d e (m+1)} \end {gather*}
Antiderivative was successfully verified.
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Rule 4723
Rule 4805
Rule 4889
Rubi steps
\begin {align*} \int (c e+d e x)^m \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac {\text {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) \text {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*}
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Mathematica [A]
time = 0.08, size = 151, normalized size = 0.83 \begin {gather*} \frac {(c+d x) (e (c+d x))^m \left ((a+b \text {ArcSin}(c+d x))^2-\frac {2 b (c+d x) (a+b \text {ArcSin}(c+d x)) \, _2F_1\left (\frac {1}{2},\frac {2+m}{2};\frac {4+m}{2};(c+d x)^2\right )}{2+m}+\frac {2 b^2 (c+d x)^2 \, _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 )}{(2+m) (3+m)}\right )}{d (1+m)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 1.38, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{m} \left (a +b \arcsin \left (d x +c \right )\right )^{2}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e \left (c + d x\right )\right )^{m} \left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{2}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,e+d\,e\,x\right )}^m\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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