Optimal. Leaf size=146 \[ \frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3}-\frac {32 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^3}-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3} \]
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Rubi [A]
time = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {45, 2392, 12,
911, 1167, 214} \begin {gather*} -\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {32 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^3}+\frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 45
Rule 214
Rule 911
Rule 1167
Rule 2392
Rubi steps
\begin {align*} \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-(b n) \int \frac {2 \left (-8 d^2-4 d e x+e^2 x^2\right )}{3 e^3 x \sqrt {d+e x}} \, dx\\ &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {(2 b n) \int \frac {-8 d^2-4 d e x+e^2 x^2}{x \sqrt {d+e x}} \, dx}{3 e^3}\\ &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {(4 b n) \text {Subst}\left (\int \frac {-3 d^2-6 d x^2+x^4}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{3 e^4}\\ &=-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {(4 b n) \text {Subst}\left (\int \left (-5 d e+e x^2-\frac {8 d^2}{-\frac {d}{e}+\frac {x^2}{e}}\right ) \, dx,x,\sqrt {d+e x}\right )}{3 e^4}\\ &=\frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3}-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (32 b d^2 n\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{3 e^4}\\ &=\frac {20 b d n \sqrt {d+e x}}{3 e^3}-\frac {4 b n (d+e x)^{3/2}}{9 e^3}-\frac {32 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^3}-\frac {2 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x}}-\frac {4 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\\ \end {align*}
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Mathematica [A]
time = 0.07, size = 124, normalized size = 0.85 \begin {gather*} \frac {-48 a d^2+56 b d^2 n-24 a d e x+52 b d e n x+6 a e^2 x^2-4 b e^2 n x^2-96 b d^{3/2} n \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )-6 b \left (8 d^2+4 d e x-e^2 x^2\right ) \log \left (c x^n\right )}{9 e^3 \sqrt {d+e x}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e x +d \right )^{\frac {3}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 159, normalized size = 1.09 \begin {gather*} \frac {4}{9} \, {\left (12 \, d^{\frac {3}{2}} e^{\left (-3\right )} \log \left (\frac {\sqrt {x e + d} - \sqrt {d}}{\sqrt {x e + d} + \sqrt {d}}\right ) - {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 15 \, \sqrt {x e + d} d\right )} e^{\left (-3\right )}\right )} b n + \frac {2}{3} \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} e^{\left (-3\right )} - 6 \, \sqrt {x e + d} d e^{\left (-3\right )} - \frac {3 \, d^{2} e^{\left (-3\right )}}{\sqrt {x e + d}}\right )} b \log \left (c x^{n}\right ) + \frac {2}{3} \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} e^{\left (-3\right )} - 6 \, \sqrt {x e + d} d e^{\left (-3\right )} - \frac {3 \, d^{2} e^{\left (-3\right )}}{\sqrt {x e + d}}\right )} a \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.41, size = 325, normalized size = 2.23 \begin {gather*} \left [\frac {2 \, {\left (24 \, {\left (b d n x e + b d^{2} n\right )} \sqrt {d} \log \left (\frac {x e - 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (28 \, b d^{2} n - {\left (2 \, b n - 3 \, a\right )} x^{2} e^{2} - 24 \, a d^{2} + 2 \, {\left (13 \, b d n - 6 \, a d\right )} x e + 3 \, {\left (b x^{2} e^{2} - 4 \, b d x e - 8 \, b d^{2}\right )} \log \left (c\right ) + 3 \, {\left (b n x^{2} e^{2} - 4 \, b d n x e - 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )}}{9 \, {\left (x e^{4} + d e^{3}\right )}}, \frac {2 \, {\left (48 \, {\left (b d n x e + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) + {\left (28 \, b d^{2} n - {\left (2 \, b n - 3 \, a\right )} x^{2} e^{2} - 24 \, a d^{2} + 2 \, {\left (13 \, b d n - 6 \, a d\right )} x e + 3 \, {\left (b x^{2} e^{2} - 4 \, b d x e - 8 \, b d^{2}\right )} \log \left (c\right ) + 3 \, {\left (b n x^{2} e^{2} - 4 \, b d n x e - 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )}}{9 \, {\left (x e^{4} + d e^{3}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 20.51, size = 262, normalized size = 1.79 \begin {gather*} \frac {- \frac {2 a d^{2}}{\sqrt {d + e x}} - 4 a d \sqrt {d + e x} + \frac {2 a \left (d + e x\right )^{\frac {3}{2}}}{3} + 2 b d^{2} \cdot \left (\frac {2 n \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} - \frac {\log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{\sqrt {d + e x}}\right ) - 4 b d \left (\sqrt {d + e x} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )} - \frac {2 n \left (\frac {d e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + e \sqrt {d + e x}\right )}{e}\right ) + 2 b \left (\frac {\left (d + e x\right )^{\frac {3}{2}} \log {\left (c \left (- \frac {d}{e} + \frac {d + e x}{e}\right )^{n} \right )}}{3} - \frac {2 n \left (\frac {d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{\sqrt {- d}} + d e \sqrt {d + e x} + \frac {e \left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{3 e}\right )}{e^{3}} \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 = -0.01 \begin {gather*} \int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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