Optimal. Leaf size=171 \[ \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}+\frac {4 b p q (f g-e h)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{5 f^{5/2} h}-\frac {4 b p q \sqrt {g+h x} (f g-e h)^2}{5 f^2 h}-\frac {4 b p q (g+h x)^{3/2} (f g-e h)}{15 f h}-\frac {4 b p q (g+h x)^{5/2}}{25 h} \]
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Rubi [A] time = 0.33, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2395, 50, 63, 208, 2445} \[ \frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\frac {4 b p q \sqrt {g+h x} (f g-e h)^2}{5 f^2 h}+\frac {4 b p q (f g-e h)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{5 f^{5/2} h}-\frac {4 b p q (g+h x)^{3/2} (f g-e h)}{15 f h}-\frac {4 b p q (g+h x)^{5/2}}{25 h} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 208
Rule 2395
Rule 2445
Rubi steps
\begin {align*} \int (g+h x)^{3/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right ) \, dx &=\operatorname {Subst}\left (\int (g+h x)^{3/2} \left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right ) \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname {Subst}\left (\frac {(2 b f p q) \int \frac {(g+h x)^{5/2}}{e+f x} \, dx}{5 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {4 b p q (g+h x)^{5/2}}{25 h}+\frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname {Subst}\left (\frac {(2 b (f g-e h) p q) \int \frac {(g+h x)^{3/2}}{e+f x} \, dx}{5 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac {4 b p q (g+h x)^{5/2}}{25 h}+\frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname {Subst}\left (\frac {\left (2 b (f g-e h)^2 p q\right ) \int \frac {\sqrt {g+h x}}{e+f x} \, dx}{5 f h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {4 b (f g-e h)^2 p q \sqrt {g+h x}}{5 f^2 h}-\frac {4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac {4 b p q (g+h x)^{5/2}}{25 h}+\frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname {Subst}\left (\frac {\left (2 b (f g-e h)^3 p q\right ) \int \frac {1}{(e+f x) \sqrt {g+h x}} \, dx}{5 f^2 h},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {4 b (f g-e h)^2 p q \sqrt {g+h x}}{5 f^2 h}-\frac {4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac {4 b p q (g+h x)^{5/2}}{25 h}+\frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}-\operatorname {Subst}\left (\frac {\left (4 b (f g-e h)^3 p q\right ) \operatorname {Subst}\left (\int \frac {1}{e-\frac {f g}{h}+\frac {f x^2}{h}} \, dx,x,\sqrt {g+h x}\right )}{5 f^2 h^2},c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=-\frac {4 b (f g-e h)^2 p q \sqrt {g+h x}}{5 f^2 h}-\frac {4 b (f g-e h) p q (g+h x)^{3/2}}{15 f h}-\frac {4 b p q (g+h x)^{5/2}}{25 h}+\frac {4 b (f g-e h)^{5/2} p q \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )}{5 f^{5/2} h}+\frac {2 (g+h x)^{5/2} \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )}{5 h}\\ \end {align*}
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Mathematica [A] time = 0.39, size = 153, normalized size = 0.89 \[ \frac {2 \left (\frac {1}{5} a (g+h x)^{5/2}+\frac {1}{5} b (g+h x)^{5/2} \log \left (c \left (d (e+f x)^p\right )^q\right )-\frac {2}{75} b p q \left (\frac {5 (f g-e h) \left (\sqrt {f} \sqrt {g+h x} (-3 e h+4 f g+f h x)-3 (f g-e h)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {g+h x}}{\sqrt {f g-e h}}\right )\right )}{f^{5/2}}+3 (g+h x)^{5/2}\right )\right )}{h} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.52, size = 624, normalized size = 3.65 \[ \left [\frac {2 \, {\left (15 \, {\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} p q \sqrt {\frac {f g - e h}{f}} \log \left (\frac {f h x + 2 \, f g - e h + 2 \, \sqrt {h x + g} f \sqrt {\frac {f g - e h}{f}}}{f x + e}\right ) + {\left (15 \, a f^{2} g^{2} - 2 \, {\left (23 \, b f^{2} g^{2} - 35 \, b e f g h + 15 \, b e^{2} h^{2}\right )} p q - 3 \, {\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \, {\left (15 \, a f^{2} g h - {\left (11 \, b f^{2} g h - 5 \, b e f h^{2}\right )} p q\right )} x + 15 \, {\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (f x + e\right ) + 15 \, {\left (b f^{2} h^{2} x^{2} + 2 \, b f^{2} g h x + b f^{2} g^{2}\right )} \log \relax (c) + 15 \, {\left (b f^{2} h^{2} q x^{2} + 2 \, b f^{2} g h q x + b f^{2} g^{2} q\right )} \log \relax (d)\right )} \sqrt {h x + g}\right )}}{75 \, f^{2} h}, \frac {2 \, {\left (30 \, {\left (b f^{2} g^{2} - 2 \, b e f g h + b e^{2} h^{2}\right )} p q \sqrt {-\frac {f g - e h}{f}} \arctan \left (-\frac {\sqrt {h x + g} f \sqrt {-\frac {f g - e h}{f}}}{f g - e h}\right ) + {\left (15 \, a f^{2} g^{2} - 2 \, {\left (23 \, b f^{2} g^{2} - 35 \, b e f g h + 15 \, b e^{2} h^{2}\right )} p q - 3 \, {\left (2 \, b f^{2} h^{2} p q - 5 \, a f^{2} h^{2}\right )} x^{2} + 2 \, {\left (15 \, a f^{2} g h - {\left (11 \, b f^{2} g h - 5 \, b e f h^{2}\right )} p q\right )} x + 15 \, {\left (b f^{2} h^{2} p q x^{2} + 2 \, b f^{2} g h p q x + b f^{2} g^{2} p q\right )} \log \left (f x + e\right ) + 15 \, {\left (b f^{2} h^{2} x^{2} + 2 \, b f^{2} g h x + b f^{2} g^{2}\right )} \log \relax (c) + 15 \, {\left (b f^{2} h^{2} q x^{2} + 2 \, b f^{2} g h q x + b f^{2} g^{2} q\right )} \log \relax (d)\right )} \sqrt {h x + g}\right )}}{75 \, f^{2} h}\right ] \]
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 \[ \int {\left (h x + g\right )}^{\frac {3}{2}} {\left (b \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.50, size = 0, normalized size = 0.00 \[ \int \left (h x +g \right )^{\frac {3}{2}} \left (b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right )+a \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 \[ \int {\left (g+h\,x\right )}^{3/2}\,\left (a+b\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 98.68, size = 484, normalized size = 2.83 \[ a g \left (\begin {cases} \sqrt {g} x & \text {for}\: h = 0 \\\frac {2 \left (g + h x\right )^{\frac {3}{2}}}{3 h} & \text {otherwise} \end {cases}\right ) + \frac {2 a \left (- \frac {g \left (g + h x\right )^{\frac {3}{2}}}{3} + \frac {\left (g + h x\right )^{\frac {5}{2}}}{5}\right )}{h} + \frac {2 b g \left (- \frac {2 f p q \left (\frac {h \left (g + h x\right )^{\frac {3}{2}}}{3 f} + \frac {\sqrt {g + h x} \left (- e h^{2} + f g h\right )}{f^{2}} + \frac {h \left (e h - f g\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {g + h x}}{\sqrt {\frac {e h - f g}{f}}} \right )}}{f^{3} \sqrt {\frac {e h - f g}{f}}}\right )}{3 h} + \frac {\left (g + h x\right )^{\frac {3}{2}} \log {\left (c \left (d \left (e - \frac {f g}{h} + \frac {f \left (g + h x\right )}{h}\right )^{p}\right )^{q} \right )}}{3}\right )}{h} + \frac {2 b \left (- \frac {2 f p q \left (\frac {h \left (g + h x\right )^{\frac {5}{2}}}{5 f} + \frac {\left (g + h x\right )^{\frac {3}{2}} \left (- e h^{2} + f g h\right )}{3 f^{2}} + \frac {\sqrt {g + h x} \left (e^{2} h^{3} - 2 e f g h^{2} + f^{2} g^{2} h\right )}{f^{3}} - \frac {h \left (e h - f g\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {g + h x}}{\sqrt {\frac {e h - f g}{f}}} \right )}}{f^{4} \sqrt {\frac {e h - f g}{f}}}\right )}{5 h} - g \left (- \frac {2 f p q \left (\frac {h \left (g + h x\right )^{\frac {3}{2}}}{3 f} + \frac {\sqrt {g + h x} \left (- e h^{2} + f g h\right )}{f^{2}} + \frac {h \left (e h - f g\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {g + h x}}{\sqrt {\frac {e h - f g}{f}}} \right )}}{f^{3} \sqrt {\frac {e h - f g}{f}}}\right )}{3 h} + \frac {\left (g + h x\right )^{\frac {3}{2}} \log {\left (c \left (d \left (e - \frac {f g}{h} + \frac {f \left (g + h x\right )}{h}\right )^{p}\right )^{q} \right )}}{3}\right ) + \frac {\left (g + h x\right )^{\frac {5}{2}} \log {\left (c \left (d \left (e - \frac {f g}{h} + \frac {f \left (g + h x\right )}{h}\right )^{p}\right )^{q} \right )}}{5}\right )}{h} \]
Verification of antiderivative is not currently implemented for this CAS.
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