Optimal. Leaf size=227 \[ -\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )\right )}{f^3 \sqrt {e+f x} (d e-c f)^3}+\frac {2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{3 f^3 (e+f x)^{3/2} (d e-c f)^2}-\frac {2 (b e-a f)^3}{5 f^3 (e+f x)^{5/2} (d e-c f)}+\frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}} \]
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Rubi [A] time = 0.28, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {87, 63, 208} \begin {gather*} -\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )\right )}{f^3 \sqrt {e+f x} (d e-c f)^3}+\frac {2 (b e-a f)^2 (a d f-3 b c f+2 b d e)}{3 f^3 (e+f x)^{3/2} (d e-c f)^2}-\frac {2 (b e-a f)^3}{5 f^3 (e+f x)^{5/2} (d e-c f)}+\frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 63
Rule 87
Rule 208
Rubi steps
\begin {align*} \int \frac {(a+b x)^3}{(c+d x) (e+f x)^{7/2}} \, dx &=\int \left (\frac {(-b e+a f)^3}{f^2 (-d e+c f) (e+f x)^{7/2}}+\frac {(-b e+a f)^2 (-2 b d e+3 b c f-a d f)}{f^2 (-d e+c f)^2 (e+f x)^{5/2}}+\frac {-3 a b^2 c^2 f^3+3 a^2 b c d f^3-a^3 d^2 f^3+b^3 e \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )}{f^2 (d e-c f)^3 (e+f x)^{3/2}}-\frac {(b c-a d)^3}{(d e-c f)^3 (c+d x) \sqrt {e+f x}}\right ) \, dx\\ &=-\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}-\frac {(b c-a d)^3 \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{(d e-c f)^3}\\ &=-\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}-\frac {\left (2 (b c-a d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{f (d e-c f)^3}\\ &=-\frac {2 (b e-a f)^3}{5 f^3 (d e-c f) (e+f x)^{5/2}}+\frac {2 (b e-a f)^2 (2 b d e-3 b c f+a d f)}{3 f^3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 (b e-a f) \left (a^2 d^2 f^2+a b d f (d e-3 c f)+b^2 \left (d^2 e^2-3 c d e f+3 c^2 f^2\right )\right )}{f^3 (d e-c f)^3 \sqrt {e+f x}}+\frac {2 (b c-a d)^3 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{\sqrt {d} (d e-c f)^{7/2}}\\ \end {align*}
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Mathematica [C] time = 0.18, size = 166, normalized size = 0.73 \begin {gather*} \frac {2 \left (-\frac {3 b \left (3 a^2 d^2 f^2-3 a b d f (c f+d e)+b^2 \left (c^2 f^2+c d e f+d^2 e^2\right )\right )}{f^3}+\frac {5 b^2 d (e+f x) (-3 a d f+b c f+2 b d e)}{f^3}+\frac {3 (b c-a d)^3 \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\frac {d (e+f x)}{d e-c f}\right )}{c f-d e}-\frac {15 b^3 d^2 (e+f x)^2}{f^3}\right )}{15 d^3 (e+f x)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.46, size = 438, normalized size = 1.93 \begin {gather*} \frac {2 (a d-b c)^3 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x} \sqrt {c f-d e}}{d e-c f}\right )}{\sqrt {d} (c f-d e)^{7/2}}-\frac {2 (a f-b e) \left (3 a^2 c^2 f^4-5 a^2 c d f^3 (e+f x)-6 a^2 c d e f^3+3 a^2 d^2 e^2 f^2+5 a^2 d^2 e f^2 (e+f x)+15 a^2 d^2 f^2 (e+f x)^2+15 a b c^2 f^3 (e+f x)-6 a b c^2 e f^3+12 a b c d e^2 f^2-20 a b c d e f^2 (e+f x)-45 a b c d f^2 (e+f x)^2-6 a b d^2 e^3 f+5 a b d^2 e^2 f (e+f x)+15 a b d^2 e f (e+f x)^2+3 b^2 c^2 e^2 f^2-15 b^2 c^2 e f^2 (e+f x)+45 b^2 c^2 f^2 (e+f x)^2-6 b^2 c d e^3 f+25 b^2 c d e^2 f (e+f x)-45 b^2 c d e f (e+f x)^2+3 b^2 d^2 e^4-10 b^2 d^2 e^3 (e+f x)+15 b^2 d^2 e^2 (e+f x)^2\right )}{15 f^3 (e+f x)^{5/2} (c f-d e)^3} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 2.09, size = 2032, normalized size = 8.95
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.36, size = 612, normalized size = 2.70 \begin {gather*} \frac {2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{{\left (c^{3} f^{3} - 3 \, c^{2} d f^{2} e + 3 \, c d^{2} f e^{2} - d^{3} e^{3}\right )} \sqrt {c d f - d^{2} e}} - \frac {2 \, {\left (45 \, {\left (f x + e\right )}^{2} a b^{2} c^{2} f^{3} - 45 \, {\left (f x + e\right )}^{2} a^{2} b c d f^{3} + 15 \, {\left (f x + e\right )}^{2} a^{3} d^{2} f^{3} + 15 \, {\left (f x + e\right )} a^{2} b c^{2} f^{4} - 5 \, {\left (f x + e\right )} a^{3} c d f^{4} + 3 \, a^{3} c^{2} f^{5} - 45 \, {\left (f x + e\right )}^{2} b^{3} c^{2} f^{2} e - 30 \, {\left (f x + e\right )} a b^{2} c^{2} f^{3} e - 15 \, {\left (f x + e\right )} a^{2} b c d f^{3} e + 5 \, {\left (f x + e\right )} a^{3} d^{2} f^{3} e - 9 \, a^{2} b c^{2} f^{4} e - 6 \, a^{3} c d f^{4} e + 45 \, {\left (f x + e\right )}^{2} b^{3} c d f e^{2} + 15 \, {\left (f x + e\right )} b^{3} c^{2} f^{2} e^{2} + 45 \, {\left (f x + e\right )} a b^{2} c d f^{2} e^{2} + 9 \, a b^{2} c^{2} f^{3} e^{2} + 18 \, a^{2} b c d f^{3} e^{2} + 3 \, a^{3} d^{2} f^{3} e^{2} - 15 \, {\left (f x + e\right )}^{2} b^{3} d^{2} e^{3} - 25 \, {\left (f x + e\right )} b^{3} c d f e^{3} - 15 \, {\left (f x + e\right )} a b^{2} d^{2} f e^{3} - 3 \, b^{3} c^{2} f^{2} e^{3} - 18 \, a b^{2} c d f^{2} e^{3} - 9 \, a^{2} b d^{2} f^{2} e^{3} + 10 \, {\left (f x + e\right )} b^{3} d^{2} e^{4} + 6 \, b^{3} c d f e^{4} + 9 \, a b^{2} d^{2} f e^{4} - 3 \, b^{3} d^{2} e^{5}\right )}}{15 \, {\left (c^{3} f^{6} - 3 \, c^{2} d f^{5} e + 3 \, c d^{2} f^{4} e^{2} - d^{3} f^{3} e^{3}\right )} {\left (f x + e\right )}^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 649, normalized size = 2.86 \begin {gather*} -\frac {2 a^{3} d^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}+\frac {6 a^{2} b c \,d^{2} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {6 a \,b^{2} c^{2} d \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}+\frac {2 b^{3} c^{3} \arctan \left (\frac {\sqrt {f x +e}\, d}{\sqrt {\left (c f -d e \right ) d}}\right )}{\left (c f -d e \right )^{3} \sqrt {\left (c f -d e \right ) d}}-\frac {2 a^{3} d^{2}}{\left (c f -d e \right )^{3} \sqrt {f x +e}}+\frac {6 a^{2} b c d}{\left (c f -d e \right )^{3} \sqrt {f x +e}}-\frac {6 a \,b^{2} c^{2}}{\left (c f -d e \right )^{3} \sqrt {f x +e}}+\frac {6 b^{3} c^{2} e}{\left (c f -d e \right )^{3} \sqrt {f x +e}\, f}-\frac {6 b^{3} c d \,e^{2}}{\left (c f -d e \right )^{3} \sqrt {f x +e}\, f^{2}}+\frac {2 b^{3} d^{2} e^{3}}{\left (c f -d e \right )^{3} \sqrt {f x +e}\, f^{3}}+\frac {2 a^{3} d}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}-\frac {2 a^{2} b c}{\left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}}}+\frac {4 a \,b^{2} c e}{\left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}} f}-\frac {2 a \,b^{2} d \,e^{2}}{\left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}} f^{2}}-\frac {2 b^{3} c \,e^{2}}{\left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}} f^{2}}+\frac {4 b^{3} d \,e^{3}}{3 \left (c f -d e \right )^{2} \left (f x +e \right )^{\frac {3}{2}} f^{3}}-\frac {2 a^{3}}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}}}+\frac {6 a^{2} b e}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}} f}-\frac {6 a \,b^{2} e^{2}}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}} f^{2}}+\frac {2 b^{3} e^{3}}{5 \left (c f -d e \right ) \left (f x +e \right )^{\frac {5}{2}} f^{3}} \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 = 1.48, size = 360, normalized size = 1.59 \begin {gather*} -\frac {\frac {2\,\left (a^3\,f^3-3\,a^2\,b\,e\,f^2+3\,a\,b^2\,e^2\,f-b^3\,e^3\right )}{5\,\left (c\,f-d\,e\right )}+\frac {2\,{\left (e+f\,x\right )}^2\,\left (a^3\,d^2\,f^3-3\,a^2\,b\,c\,d\,f^3+3\,a\,b^2\,c^2\,f^3-3\,b^3\,c^2\,e\,f^2+3\,b^3\,c\,d\,e^2\,f-b^3\,d^2\,e^3\right )}{{\left (c\,f-d\,e\right )}^3}-\frac {2\,\left (e+f\,x\right )\,\left (d\,a^3\,f^3-3\,c\,a^2\,b\,f^3-3\,d\,a\,b^2\,e^2\,f+6\,c\,a\,b^2\,e\,f^2+2\,d\,b^3\,e^3-3\,c\,b^3\,e^2\,f\right )}{3\,{\left (c\,f-d\,e\right )}^2}}{f^3\,{\left (e+f\,x\right )}^{5/2}}-\frac {2\,\mathrm {atan}\left (\frac {2\,\sqrt {d}\,\sqrt {e+f\,x}\,{\left (a\,d-b\,c\right )}^3\,\left (c^3\,f^3-3\,c^2\,d\,e\,f^2+3\,c\,d^2\,e^2\,f-d^3\,e^3\right )}{{\left (c\,f-d\,e\right )}^{7/2}\,\left (2\,a^3\,d^3-6\,a^2\,b\,c\,d^2+6\,a\,b^2\,c^2\,d-2\,b^3\,c^3\right )}\right )\,{\left (a\,d-b\,c\right )}^3}{\sqrt {d}\,{\left (c\,f-d\,e\right )}^{7/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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