Optimal. Leaf size=472 \[ \frac {i b f^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 i b f^5 \left (1+c^2 x^2\right )^{5/2}}{3 c (i-c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 b f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^5 (1-i c x)^4 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {10 i f^5 (1-i c x)^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 i f^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 b f^5 \left (1+c^2 x^2\right )^{5/2} \log (i-c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
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Rubi [A]
time = 0.30, antiderivative size = 472, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {5796, 683,
655, 221, 5837, 641, 45, 5783} \begin {gather*} -\frac {5 i f^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {10 i f^5 (1-i c x)^2 \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^5 (1-i c x)^4 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 f^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b f^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 i b f^5 \left (c^2 x^2+1\right )^{5/2}}{3 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 b f^5 \left (c^2 x^2+1\right )^{5/2} \log (-c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 b f^5 \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 221
Rule 641
Rule 655
Rule 683
Rule 5783
Rule 5796
Rule 5837
Rubi steps
\begin {align*} \int \frac {(f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{(d+i c d x)^{5/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(f-i c f x)^5 \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {2 i f^5 (1-i c x)^4 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {10 i f^5 (1-i c x)^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 i f^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {5 i f^5}{c}+\frac {2 i f^5 (1-i c x)^4}{3 c \left (1+c^2 x^2\right )^2}-\frac {10 i f^5 (1-i c x)^2}{3 c \left (1+c^2 x^2\right )}+\frac {5 f^5 \sinh ^{-1}(c x)}{c \sqrt {1+c^2 x^2}}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {5 i b f^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^5 (1-i c x)^4 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {10 i f^5 (1-i c x)^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 i f^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {(1-i c x)^4}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (10 i b f^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {(1-i c x)^2}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (5 b f^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {5 i b f^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 b f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^5 (1-i c x)^4 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {10 i f^5 (1-i c x)^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 i f^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {(1-i c x)^2}{(1+i c x)^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (10 i b f^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1-i c x}{1+i c x} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {5 i b f^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 b f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^5 (1-i c x)^4 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {10 i f^5 (1-i c x)^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 i f^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (1-\frac {4}{(-i+c x)^2}+\frac {4 i}{-i+c x}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (10 i b f^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-1-\frac {2 i}{-i+c x}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {i b f^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 i b f^5 \left (1+c^2 x^2\right )^{5/2}}{3 c (i-c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 b f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^5 (1-i c x)^4 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {10 i f^5 (1-i c x)^2 \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 i f^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 f^5 \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 b f^5 \left (1+c^2 x^2\right )^{5/2} \log (i-c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(1005\) vs. \(2(472)=944\).
time = 4.91, size = 1005, normalized size = 2.13 \begin {gather*} \frac {-\frac {4 i a f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (-23-34 i c x+3 c^2 x^2\right )}{d^3 (-i+c x)^2}+\frac {60 a f^{5/2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )}{d^{5/2}}-\frac {2 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (\cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right ) \left (\left (-14+3 i \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)-28 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+7 i \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (84 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-i \left (8-6 i \sinh ^{-1}(c x)+9 \sinh ^{-1}(c x)^2+21 \log \left (1+c^2 x^2\right )\right )\right )+2 \left (4-4 i \sinh ^{-1}(c x)+6 \sinh ^{-1}(c x)^2+56 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+14 \log \left (1+c^2 x^2\right )+\sqrt {1+c^2 x^2} \left (\sinh ^{-1}(c x) \left (-14 i+3 \sinh ^{-1}(c x)\right )+28 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+7 \log \left (1+c^2 x^2\right )\right )\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{d^3 (i+c x) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^4}+\frac {2 i b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (-i \cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right ) \left (\sinh ^{-1}(c x)-2 \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-\frac {1}{2} i \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right ) \left (4+3 i \sinh ^{-1}(c x)-6 i \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\frac {3}{2} \log \left (1+c^2 x^2\right )\right )+2 \left (\left (2+\sqrt {1+c^2 x^2}\right ) \sinh ^{-1}(c x)+2 \left (2+\sqrt {1+c^2 x^2}\right ) \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\frac {1}{2} i \left (4+\left (2+\sqrt {1+c^2 x^2}\right ) \log \left (1+c^2 x^2\right )\right )\right ) \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{d^3 (i+c x) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^4}+\frac {b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (2 \left (4+6 i c x-6 c^2 x^2+52 (-i+c x) \text {ArcTan}\left (\coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+13 (1+i c x) \log \left (1+c^2 x^2\right )\right ) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+18 \sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^3+\sinh ^{-1}(c x) \left (-24 i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-35 i \cosh \left (\frac {3}{2} \sinh ^{-1}(c x)\right )+3 i \cosh \left (\frac {5}{2} \sinh ^{-1}(c x)\right )-24 \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+35 \sinh \left (\frac {3}{2} \sinh ^{-1}(c x)\right )+3 \sinh \left (\frac {5}{2} \sinh ^{-1}(c x)\right )\right )\right )}{d^3 (i+c x) \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )^4}}{12 c} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (-i c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )}{\left (i c d x +d \right )^{\frac {5}{2}}}\, dx\]
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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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 = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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