Optimal. Leaf size=261 \[ \frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac {6 b e^{3/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {(c+d x)^2+1}}-\frac {12 b e^{3/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {(c+d x)^2+1}}-\frac {4 b \sqrt {(c+d x)^2+1} (e (c+d x))^{3/2}}{25 d}+\frac {12 b e \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{25 d (c+d x+1)} \]
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Rubi [A] time = 0.24, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5865, 5661, 321, 329, 305, 220, 1196} \[ \frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac {6 b e^{3/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {(c+d x)^2+1}}-\frac {12 b e^{3/2} (c+d x+1) \sqrt {\frac {(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {(c+d x)^2+1}}-\frac {4 b \sqrt {(c+d x)^2+1} (e (c+d x))^{3/2}}{25 d}+\frac {12 b e \sqrt {(c+d x)^2+1} \sqrt {e (c+d x)}}{25 d (c+d x+1)} \]
Antiderivative was successfully verified.
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Rule 220
Rule 305
Rule 321
Rule 329
Rule 1196
Rule 5661
Rule 5865
Rubi steps
\begin {align*} \int (c e+d e x)^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int (e x)^{3/2} \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {(e x)^{5/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=-\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac {(6 b e) \operatorname {Subst}\left (\int \frac {\sqrt {e x}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{25 d}\\ &=-\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac {(12 b) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{25 d}\\ &=-\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}+\frac {(12 b e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{25 d}-\frac {(12 b e) \operatorname {Subst}\left (\int \frac {1-\frac {x^2}{e}}{\sqrt {1+\frac {x^4}{e^2}}} \, dx,x,\sqrt {e (c+d x)}\right )}{25 d}\\ &=-\frac {4 b (e (c+d x))^{3/2} \sqrt {1+(c+d x)^2}}{25 d}+\frac {12 b e \sqrt {e (c+d x)} \sqrt {1+(c+d x)^2}}{25 d (1+c+d x)}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e}-\frac {12 b e^{3/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {1+(c+d x)^2}}+\frac {6 b e^{3/2} (1+c+d x) \sqrt {\frac {1+(c+d x)^2}{(1+c+d x)^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt {e (c+d x)}}{\sqrt {e}}\right )|\frac {1}{2}\right )}{25 d \sqrt {1+(c+d x)^2}}\\ \end {align*}
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Mathematica [C] time = 0.05, size = 87, normalized size = 0.33 \[ \frac {2 (e (c+d x))^{3/2} \left (5 a c+5 a d x+2 b \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-(c+d x)^2\right )-2 b \sqrt {(c+d x)^2+1}+5 b c \sinh ^{-1}(c+d x)+5 b d x \sinh ^{-1}(c+d x)\right )}{25 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a d e x + a c e + {\left (b d e x + b c e\right )} \operatorname {arsinh}\left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]
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 \[ \int {\left (d e x + c e\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.01, size = 205, normalized size = 0.79 \[ \frac {\frac {2 \left (d e x +c e \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d e x +c e \right )^{\frac {5}{2}} \arcsinh \left (\frac {d e x +c e}{e}\right )}{5}-\frac {2 \left (\frac {e^{2} \left (d e x +c e \right )^{\frac {3}{2}} \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{5}-\frac {3 i e^{3} \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{5 \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}\right )}{5 e}\right )}{d e} \]
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 \[ \frac {1}{50} \, {\left (\frac {20 \, {\left (d^{2} e^{\frac {3}{2}} x^{2} + 2 \, c d e^{\frac {3}{2}} x + c^{2} e^{\frac {3}{2}}\right )} \sqrt {d x + c} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d} - \frac {8 \, {\left (d x + c\right )}^{\frac {5}{2}} e^{\frac {3}{2}} - 40 \, \sqrt {d x + c} e^{\frac {3}{2}} - 5 \, {\left (i \, \sqrt {2} e {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - i \, \sqrt {2} e {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - \sqrt {2} e \log \left (d x + \sqrt {2} \sqrt {d x + c} + c + 1\right ) + \sqrt {2} e \log \left (d x - \sqrt {2} \sqrt {d x + c} + c + 1\right )\right )} \sqrt {e}}{d} - 50 \, \int \frac {2 \, {\left (d^{2} e^{\frac {3}{2}} x^{2} + 2 \, c d e^{\frac {3}{2}} x + c^{2} e^{\frac {3}{2}}\right )} \sqrt {d x + c}}{5 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c\right )}}\,{d x}\right )} b + \frac {2 \, {\left (d e x + c e\right )}^{\frac {5}{2}} a}{5 \, d e} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,e+d\,e\,x\right )}^{3/2}\,\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right ) \,d x \]
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 \[ \int \left (e \left (c + d x\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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