Optimal. Leaf size=112 \[ -\frac {2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {2 \sqrt {d+e x} (b d-a e)^2}{b^3}+\frac {2 (d+e x)^{3/2} (b d-a e)}{3 b^2}+\frac {2 (d+e x)^{5/2}}{5 b} \]
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Rubi [A] time = 0.06, antiderivative size = 112, 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, 50, 63, 208} \begin {gather*} \frac {2 \sqrt {d+e x} (b d-a e)^2}{b^3}+\frac {2 (d+e x)^{3/2} (b d-a e)}{3 b^2}-\frac {2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {2 (d+e x)^{5/2}}{5 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 50
Rule 63
Rule 208
Rubi steps
\begin {align*} \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac {(d+e x)^{5/2}}{a+b x} \, dx\\ &=\frac {2 (d+e x)^{5/2}}{5 b}+\frac {(b d-a e) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{b}\\ &=\frac {2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac {2 (d+e x)^{5/2}}{5 b}+\frac {(b d-a e)^2 \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{b^2}\\ &=\frac {2 (b d-a e)^2 \sqrt {d+e x}}{b^3}+\frac {2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac {2 (d+e x)^{5/2}}{5 b}+\frac {(b d-a e)^3 \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{b^3}\\ &=\frac {2 (b d-a e)^2 \sqrt {d+e x}}{b^3}+\frac {2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac {2 (d+e x)^{5/2}}{5 b}+\frac {\left (2 (b d-a e)^3\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b^3 e}\\ &=\frac {2 (b d-a e)^2 \sqrt {d+e x}}{b^3}+\frac {2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac {2 (d+e x)^{5/2}}{5 b}-\frac {2 (b d-a e)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 105, normalized size = 0.94 \begin {gather*} \frac {2 (b d-a e) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )}{3 b^{7/2}}+\frac {2 (d+e x)^{5/2}}{5 b} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.12, size = 130, normalized size = 1.16 \begin {gather*} \frac {2 \sqrt {d+e x} \left (15 a^2 e^2-5 a b e (d+e x)-30 a b d e+15 b^2 d^2+3 b^2 (d+e x)^2+5 b^2 d (d+e x)\right )}{15 b^3}+\frac {2 (a e-b d)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 290, normalized size = 2.59 \begin {gather*} \left [\frac {15 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 180, normalized size = 1.61 \begin {gather*} \frac {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 )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d + 15 \, \sqrt {x e + d} b^{4} d^{2} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} e - 30 \, \sqrt {x e + d} a b^{3} d e + 15 \, \sqrt {x e + d} a^{2} b^{2} e^{2}\right )}}{15 \, b^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 263, normalized size = 2.35 \begin {gather*} -\frac {2 a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {6 a^{2} d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {6 a \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 d^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}+\frac {2 \sqrt {e x +d}\, a^{2} e^{2}}{b^{3}}-\frac {4 \sqrt {e x +d}\, a d e}{b^{2}}+\frac {2 \sqrt {e x +d}\, d^{2}}{b}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} a e}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} d}{3 b}+\frac {2 \left (e x +d \right )^{\frac {5}{2}}}{5 b} \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 = 0.07, size = 130, normalized size = 1.16 \begin {gather*} \frac {2\,{\left (d+e\,x\right )}^{5/2}}{5\,b}-\frac {2\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}+\frac {2\,{\left (a\,e-b\,d\right )}^2\,\sqrt {d+e\,x}}{b^3}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{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 )}^{5/2}}{b^{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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