Optimal. Leaf size=130 \[ \frac {2 (b c-a d) (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}}-\frac {2 \sqrt {e+f x} (b c-a d) (d e-c f)}{d^3}-\frac {2 (e+f x)^{3/2} (b c-a d)}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f} \]
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Rubi [A] time = 0.11, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \begin {gather*} -\frac {2 (e+f x)^{3/2} (b c-a d)}{3 d^2}-\frac {2 \sqrt {e+f x} (b c-a d) (d e-c f)}{d^3}+\frac {2 (b c-a d) (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}}+\frac {2 b (e+f x)^{5/2}}{5 d f} \end {gather*}
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 80
Rule 208
Rubi steps
\begin {align*} \int \frac {(a+b x) (e+f x)^{3/2}}{c+d x} \, dx &=\frac {2 b (e+f x)^{5/2}}{5 d f}+\frac {\left (2 \left (-\frac {5}{2} b c f+\frac {5 a d f}{2}\right )\right ) \int \frac {(e+f x)^{3/2}}{c+d x} \, dx}{5 d f}\\ &=-\frac {2 (b c-a d) (e+f x)^{3/2}}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {((b c-a d) (d e-c f)) \int \frac {\sqrt {e+f x}}{c+d x} \, dx}{d^2}\\ &=-\frac {2 (b c-a d) (d e-c f) \sqrt {e+f x}}{d^3}-\frac {2 (b c-a d) (e+f x)^{3/2}}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {\left ((b c-a d) (d e-c f)^2\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{d^3}\\ &=-\frac {2 (b c-a d) (d e-c f) \sqrt {e+f x}}{d^3}-\frac {2 (b c-a d) (e+f x)^{3/2}}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f}-\frac {\left (2 (b c-a d) (d e-c f)^2\right ) \operatorname {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{d^3 f}\\ &=-\frac {2 (b c-a d) (d e-c f) \sqrt {e+f x}}{d^3}-\frac {2 (b c-a d) (e+f x)^{3/2}}{3 d^2}+\frac {2 b (e+f x)^{5/2}}{5 d f}+\frac {2 (b c-a d) (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{d^{7/2}}\\ \end {align*}
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Mathematica [A] time = 0.21, size = 112, normalized size = 0.86 \begin {gather*} \frac {2 \left (\frac {5 (a d f-b c f) \left (\sqrt {d} \sqrt {e+f x} (-3 c f+4 d e+d f x)-3 (d e-c f)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )\right )}{3 d^{5/2}}+b (e+f x)^{5/2}\right )}{5 d f} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.20, size = 154, normalized size = 1.18 \begin {gather*} \frac {2 \sqrt {e+f x} \left (-15 a c d f^2+5 a d^2 f (e+f x)+15 a d^2 e f+15 b c^2 f^2-5 b c d f (e+f x)-15 b c d e f+3 b d^2 (e+f x)^2\right )}{15 d^3 f}-\frac {2 (a d-b c) (c f-d e)^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{d^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.35, size = 373, normalized size = 2.87 \begin {gather*} \left [-\frac {15 \, {\left ({\left (b c d - a d^{2}\right )} e f - {\left (b c^{2} - a c d\right )} f^{2}\right )} \sqrt {\frac {d e - c f}{d}} \log \left (\frac {d f x + 2 \, d e - c f - 2 \, \sqrt {f x + e} d \sqrt {\frac {d e - c f}{d}}}{d x + c}\right ) - 2 \, {\left (3 \, b d^{2} f^{2} x^{2} + 3 \, b d^{2} e^{2} - 20 \, {\left (b c d - a d^{2}\right )} e f + 15 \, {\left (b c^{2} - a c d\right )} f^{2} + {\left (6 \, b d^{2} e f - 5 \, {\left (b c d - a d^{2}\right )} f^{2}\right )} x\right )} \sqrt {f x + e}}{15 \, d^{3} f}, \frac {2 \, {\left (15 \, {\left ({\left (b c d - a d^{2}\right )} e f - {\left (b c^{2} - a c d\right )} f^{2}\right )} \sqrt {-\frac {d e - c f}{d}} \arctan \left (-\frac {\sqrt {f x + e} d \sqrt {-\frac {d e - c f}{d}}}{d e - c f}\right ) + {\left (3 \, b d^{2} f^{2} x^{2} + 3 \, b d^{2} e^{2} - 20 \, {\left (b c d - a d^{2}\right )} e f + 15 \, {\left (b c^{2} - a c d\right )} f^{2} + {\left (6 \, b d^{2} e f - 5 \, {\left (b c d - a d^{2}\right )} f^{2}\right )} x\right )} \sqrt {f x + e}\right )}}{15 \, d^{3} f}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.21, size = 238, normalized size = 1.83 \begin {gather*} -\frac {2 \, {\left (b c^{3} f^{2} - a c^{2} d f^{2} - 2 \, b c^{2} d f e + 2 \, a c d^{2} f e + b c d^{2} e^{2} - a d^{3} e^{2}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{\sqrt {c d f - d^{2} e} d^{3}} + \frac {2 \, {\left (3 \, {\left (f x + e\right )}^{\frac {5}{2}} b d^{4} f^{4} - 5 \, {\left (f x + e\right )}^{\frac {3}{2}} b c d^{3} f^{5} + 5 \, {\left (f x + e\right )}^{\frac {3}{2}} a d^{4} f^{5} + 15 \, \sqrt {f x + e} b c^{2} d^{2} f^{6} - 15 \, \sqrt {f x + e} a c d^{3} f^{6} - 15 \, \sqrt {f x + e} b c d^{3} f^{5} e + 15 \, \sqrt {f x + e} a d^{4} f^{5} e\right )}}{15 \, d^{5} f^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 370, normalized size = 2.85 \begin {gather*} \frac {2 a \,c^{2} f^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{2}}-\frac {4 a c e f \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d}+\frac {2 a \,e^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}}-\frac {2 b \,c^{3} f^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{3}}+\frac {4 b \,c^{2} e f \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d^{2}}-\frac {2 b c \,e^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\sqrt {\left (c f -d e \right ) d}\, d}-\frac {2 \sqrt {f x +e}\, a c f}{d^{2}}+\frac {2 \sqrt {f x +e}\, a e}{d}+\frac {2 \sqrt {f x +e}\, b \,c^{2} f}{d^{3}}-\frac {2 \sqrt {f x +e}\, b c e}{d^{2}}+\frac {2 \left (f x +e \right )^{\frac {3}{2}} a}{3 d}-\frac {2 \left (f x +e \right )^{\frac {3}{2}} b c}{3 d^{2}}+\frac {2 \left (f x +e \right )^{\frac {5}{2}} b}{5 d f} \end {gather*}
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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.23, size = 236, normalized size = 1.82 \begin {gather*} {\left (e+f\,x\right )}^{3/2}\,\left (\frac {2\,a\,f-2\,b\,e}{3\,d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{3\,d^2\,f^2}\right )+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{3/2}}{-b\,c^3\,f^2+2\,b\,c^2\,d\,e\,f+a\,c^2\,d\,f^2-b\,c\,d^2\,e^2-2\,a\,c\,d^2\,e\,f+a\,d^3\,e^2}\right )\,\left (a\,d-b\,c\right )\,{\left (c\,f-d\,e\right )}^{3/2}}{d^{7/2}}+\frac {2\,b\,{\left (e+f\,x\right )}^{5/2}}{5\,d\,f}-\frac {\sqrt {e+f\,x}\,\left (c\,f^2-d\,e\,f\right )\,\left (\frac {2\,a\,f-2\,b\,e}{d\,f}-\frac {2\,b\,\left (c\,f^2-d\,e\,f\right )}{d^2\,f^2}\right )}{d\,f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 40.66, size = 139, normalized size = 1.07 \begin {gather*} \frac {2 b \left (e + f x\right )^{\frac {5}{2}}}{5 d f} + \frac {\left (e + f x\right )^{\frac {3}{2}} \left (2 a d - 2 b c\right )}{3 d^{2}} + \frac {\sqrt {e + f x} \left (- 2 a c d f + 2 a d^{2} e + 2 b c^{2} f - 2 b c d e\right )}{d^{3}} + \frac {2 \left (a d - b c\right ) \left (c f - d e\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{d^{4} \sqrt {\frac {c f - d e}{d}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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