Optimal. Leaf size=295 \[ \frac{f^2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 i f^2 (1-i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b f^2 \left (c^2 x^2+1\right )^{5/2}}{3 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b f^2 \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b f^2 \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
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Rubi [A] time = 0.337132, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {5712, 653, 191, 5819, 627, 44, 203, 260} \[ \frac{f^2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 i f^2 (1-i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b f^2 \left (c^2 x^2+1\right )^{5/2}}{3 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b f^2 \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b f^2 \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 5712
Rule 653
Rule 191
Rule 5819
Rule 627
Rule 44
Rule 203
Rule 260
Rubi steps
\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} \sqrt{f-i c f x}} \, dx &=\frac{\left (1+c^2 x^2\right )^{5/2} \int \frac{(f-i c f x)^2 \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^2 (1-i c x) \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{f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (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{2 i f^2 (1-i c x)}{3 c \left (1+c^2 x^2\right )^2}+\frac{f^2 x}{3 \left (1+c^2 x^2\right )}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{2 i f^2 (1-i c x) \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{f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1-i c x}{\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 (b c f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{2 i f^2 (1-i c x) \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{f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1}{(1-i c x) (1+i c x)^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{2 i f^2 (1-i c x) \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{f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac{1}{2 (-i+c x)^2}+\frac{1}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{i b f^2 \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{2 i f^2 (1-i c x) \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{f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{i b f^2 \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{2 i f^2 (1-i c x) \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{f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b f^2 \left (1+c^2 x^2\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.495634, size = 143, normalized size = 0.48 \[ \frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left ((c x-2 i) \left (a \sqrt{c^2 x^2+1}+b c x-i b\right )+b (c x-2 i) \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)-b (c x-i)^2 \log (d+i c d x)\right )}{3 c d^3 f (c x-i)^2 \sqrt{c^2 x^2+1}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.26, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( d+icdx \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{f-icfx}}}}\, 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 [B] time = 2.91807, size = 1386, normalized size = 4.7 \begin{align*} -\frac{12 \, \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b c x - 12 \,{\left (b c^{2} x^{2} - i \, b c x + 2 \, b\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 2 \,{\left (3 \, c^{4} d^{3} f x^{3} - 3 i \, c^{3} d^{3} f x^{2} + 3 \, c^{2} d^{3} f x - 3 i \, c d^{3} f\right )} \sqrt{\frac{b^{2}}{c^{2} d^{5} f}} \log \left (\frac{3 \,{\left (-2 i \, b c^{6} x^{2} - 4 \, b c^{5} x + 4 i \, b c^{4}\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} + 2 \,{\left (3 i \, c^{9} d^{3} f x^{4} + 6 \, c^{8} d^{3} f x^{3} + 3 i \, c^{7} d^{3} f x^{2} + 6 \, c^{6} d^{3} f x\right )} \sqrt{\frac{b^{2}}{c^{2} d^{5} f}}}{3 \,{\left (16 \, b c^{3} x^{3} - 16 i \, b c^{2} x^{2} + 16 \, b c x - 16 i \, b\right )}}\right ) + 2 \,{\left (3 \, c^{4} d^{3} f x^{3} - 3 i \, c^{3} d^{3} f x^{2} + 3 \, c^{2} d^{3} f x - 3 i \, c d^{3} f\right )} \sqrt{\frac{b^{2}}{c^{2} d^{5} f}} \log \left (\frac{3 \,{\left (-2 i \, b c^{6} x^{2} - 4 \, b c^{5} x + 4 i \, b c^{4}\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} + 2 \,{\left (-3 i \, c^{9} d^{3} f x^{4} - 6 \, c^{8} d^{3} f x^{3} - 3 i \, c^{7} d^{3} f x^{2} - 6 \, c^{6} d^{3} f x\right )} \sqrt{\frac{b^{2}}{c^{2} d^{5} f}}}{3 \,{\left (16 \, b c^{3} x^{3} - 16 i \, b c^{2} x^{2} + 16 \, b c x - 16 i \, b\right )}}\right ) - 3 \,{\left (4 \, a c^{2} x^{2} - 4 i \, a c x + 8 \, a\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}}{3 \,{\left (12 \, c^{4} d^{3} f x^{3} - 12 i \, c^{3} d^{3} f x^{2} + 12 \, c^{2} d^{3} f x - 12 i \, c d^{3} f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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