Optimal. Leaf size=140 \[ \frac {15 e^2}{4 \sqrt {d+e x} (b d-a e)^3}-\frac {15 \sqrt {b} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{7/2}}+\frac {5 e}{4 (a+b x) \sqrt {d+e x} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)} \]
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Rubi [A] time = 0.06, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \begin {gather*} \frac {15 e^2}{4 \sqrt {d+e x} (b d-a e)^3}-\frac {15 \sqrt {b} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{7/2}}+\frac {5 e}{4 (a+b x) \sqrt {d+e x} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 51
Rule 63
Rule 208
Rubi steps
\begin {align*} \int \frac {a+b x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a+b x)^3 (d+e x)^{3/2}} \, dx\\ &=-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {(5 e) \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}} \, dx}{4 (b d-a e)}\\ &=-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {\left (15 e^2\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^2}\\ &=\frac {15 e^2}{4 (b d-a e)^3 \sqrt {d+e x}}-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {\left (15 b e^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 (b d-a e)^3}\\ &=\frac {15 e^2}{4 (b d-a e)^3 \sqrt {d+e x}}-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {(15 b e) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 (b d-a e)^3}\\ &=\frac {15 e^2}{4 (b d-a e)^3 \sqrt {d+e x}}-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {15 \sqrt {b} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{7/2}}\\ \end {align*}
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Mathematica [C] time = 0.01, size = 50, normalized size = 0.36 \begin {gather*} -\frac {2 e^2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{\sqrt {d+e x} (a e-b d)^3} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.54, size = 163, normalized size = 1.16 \begin {gather*} \frac {e^2 \left (8 a^2 e^2+25 a b e (d+e x)-16 a b d e+8 b^2 d^2+15 b^2 (d+e x)^2-25 b^2 d (d+e x)\right )}{4 \sqrt {d+e x} (b d-a e)^3 (-a e-b (d+e x)+b d)^2}+\frac {15 \sqrt {b} e^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 (a e-b d)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.45, size = 782, normalized size = 5.59 \begin {gather*} \left [-\frac {15 \, {\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} + {\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} + {\left (2 \, a b d e^{2} + a^{2} e^{3}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \, {\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} + {\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} + {\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} + {\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}, -\frac {15 \, {\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} + {\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} + {\left (2 \, a b d e^{2} + a^{2} e^{3}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (15 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \, {\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} + {\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} + {\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} + {\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 235, normalized size = 1.68 \begin {gather*} \frac {15 \, b \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, e^{2}}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {x e + d}} + \frac {7 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} e^{2} - 9 \, \sqrt {x e + d} b^{2} d e^{2} + 9 \, \sqrt {x e + d} a b e^{3}}{4 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 179, normalized size = 1.28 \begin {gather*} -\frac {9 \sqrt {e x +d}\, a b \,e^{3}}{4 \left (a e -b d \right )^{3} \left (b e x +a e \right )^{2}}+\frac {9 \sqrt {e x +d}\, b^{2} d \,e^{2}}{4 \left (a e -b d \right )^{3} \left (b e x +a e \right )^{2}}-\frac {7 \left (e x +d \right )^{\frac {3}{2}} b^{2} e^{2}}{4 \left (a e -b d \right )^{3} \left (b e x +a e \right )^{2}}-\frac {15 b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}-\frac {2 e^{2}}{\left (a e -b d \right )^{3} \sqrt {e x +d}} \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 = 2.32, size = 205, normalized size = 1.46 \begin {gather*} -\frac {\frac {2\,e^2}{a\,e-b\,d}+\frac {15\,b^2\,e^2\,{\left (d+e\,x\right )}^2}{4\,{\left (a\,e-b\,d\right )}^3}+\frac {25\,b\,e^2\,\left (d+e\,x\right )}{4\,{\left (a\,e-b\,d\right )}^2}}{b^2\,{\left (d+e\,x\right )}^{5/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{3/2}+\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}-\frac {15\,\sqrt {b}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}}\right )}{4\,{\left (a\,e-b\,d\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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