Optimal. Leaf size=128 \[ \frac {16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )}{15 d e^3}-\frac {8 b (e (c+d x))^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^2}+\frac {2 \sqrt {e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e} \]
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Rubi [A] time = 0.19, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4805, 4627, 4711} \[ \frac {16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )}{15 d e^3}-\frac {8 b (e (c+d x))^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{3 d e^2}+\frac {2 \sqrt {e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e} \]
Antiderivative was successfully verified.
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Rule 4627
Rule 4711
Rule 4805
Rubi steps
\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{\sqrt {c e+d e x}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{\sqrt {e x}} \, dx,x,c+d x\right )}{d}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {\sqrt {e x} \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e}\\ &=\frac {2 \sqrt {e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}-\frac {8 b (e (c+d x))^{3/2} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right )}{3 d e^2}+\frac {16 b^2 (e (c+d x))^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )}{15 d e^3}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 107, normalized size = 0.84 \[ \frac {2 \sqrt {e (c+d x)} \left (8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};(c+d x)^2\right )+5 \left (a+b \sin ^{-1}(c+d x)\right ) \left (3 \left (a+b \sin ^{-1}(c+d x)\right )-4 b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};(c+d x)^2\right )\right )\right )}{15 d e} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}{\sqrt {d e x + c e}}, x\right ) \]
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 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}}{\sqrt {d e x + c e}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F(-2)] time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (d x +c \right )\right )^{2}}{\sqrt {d e x +c e}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2}{\sqrt {c\,e+d\,e\,x}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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