Optimal. Leaf size=219 \[ -\frac {15 b^{5/2} e^{-\frac {a}{b m n}} m^{5/2} n^{5/2} \sqrt {\pi } (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{\sqrt {b} \sqrt {m} \sqrt {n}}\right )}{8 f}+\frac {15 b^2 m^2 n^2 (e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{4 f}-\frac {5 b m n (e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}{2 f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f} \]
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Rubi [A]
time = 0.26, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2436, 2333,
2337, 2211, 2235, 2495} \begin {gather*} -\frac {15 \sqrt {\pi } b^{5/2} m^{5/2} n^{5/2} (e+f x) e^{-\frac {a}{b m n}} \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {Erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{\sqrt {b} \sqrt {m} \sqrt {n}}\right )}{8 f}+\frac {15 b^2 m^2 n^2 (e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{4 f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f}-\frac {5 b m n (e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}{2 f} \end {gather*}
Antiderivative was successfully verified.
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Rule 2211
Rule 2235
Rule 2333
Rule 2337
Rule 2436
Rule 2495
Rubi steps
\begin {align*} \int \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2} \, dx &=\text {Subst}\left (\int \left (a+b \log \left (c d^n (e+f x)^{m n}\right )\right )^{5/2} \, dx,c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right )\\ &=\text {Subst}\left (\frac {\text {Subst}\left (\int \left (a+b \log \left (c d^n x^{m n}\right )\right )^{5/2} \, dx,x,e+f x\right )}{f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right )\\ &=\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f}-\text {Subst}\left (\frac {(5 b m n) \text {Subst}\left (\int \left (a+b \log \left (c d^n x^{m n}\right )\right )^{3/2} \, dx,x,e+f x\right )}{2 f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right )\\ &=-\frac {5 b m n (e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}{2 f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f}+\text {Subst}\left (\frac {\left (15 b^2 m^2 n^2\right ) \text {Subst}\left (\int \sqrt {a+b \log \left (c d^n x^{m n}\right )} \, dx,x,e+f x\right )}{4 f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right )\\ &=\frac {15 b^2 m^2 n^2 (e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{4 f}-\frac {5 b m n (e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}{2 f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f}-\text {Subst}\left (\frac {\left (15 b^3 m^3 n^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b \log \left (c d^n x^{m n}\right )}} \, dx,x,e+f x\right )}{8 f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right )\\ &=\frac {15 b^2 m^2 n^2 (e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{4 f}-\frac {5 b m n (e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}{2 f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f}-\text {Subst}\left (\frac {\left (15 b^3 m^2 n^2 (e+f x) \left (c d^n (e+f x)^{m n}\right )^{-\frac {1}{m n}}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{m n}}}{\sqrt {a+b x}} \, dx,x,\log \left (c d^n (e+f x)^{m n}\right )\right )}{8 f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right )\\ &=\frac {15 b^2 m^2 n^2 (e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{4 f}-\frac {5 b m n (e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}{2 f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f}-\text {Subst}\left (\frac {\left (15 b^2 m^2 n^2 (e+f x) \left (c d^n (e+f x)^{m n}\right )^{-\frac {1}{m n}}\right ) \text {Subst}\left (\int e^{-\frac {a}{b m n}+\frac {x^2}{b m n}} \, dx,x,\sqrt {a+b \log \left (c d^n (e+f x)^{m n}\right )}\right )}{4 f},c d^n (e+f x)^{m n},c \left (d (e+f x)^m\right )^n\right )\\ &=-\frac {15 b^{5/2} e^{-\frac {a}{b m n}} m^{5/2} n^{5/2} \sqrt {\pi } (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{\sqrt {b} \sqrt {m} \sqrt {n}}\right )}{8 f}+\frac {15 b^2 m^2 n^2 (e+f x) \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{4 f}-\frac {5 b m n (e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}{2 f}+\frac {(e+f x) \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}}{f}\\ \end {align*}
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Mathematica [A]
time = 0.22, size = 190, normalized size = 0.87 \begin {gather*} \frac {(e+f x) \left (8 \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{5/2}-5 b m n \left (3 b^{3/2} e^{-\frac {a}{b m n}} m^{3/2} n^{3/2} \sqrt {\pi } \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{\sqrt {b} \sqrt {m} \sqrt {n}}\right )+2 \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )} \left (2 a-3 b m n+2 b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )\right )\right )}{8 f} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \left (a +b \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )\right )^{\frac {5}{2}}\, dx\]
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(-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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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]
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 = -0.00 \begin {gather*} \int {\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^m\right )}^n\right )\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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