Optimal. Leaf size=241 \[ -\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^3 (a+b x) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.14, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {768, 646, 47, 50, 63, 208} \begin {gather*} -\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^3 (a+b x) \sqrt {b d-a e} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 47
Rule 50
Rule 63
Rule 208
Rule 646
Rule 768
Rubi steps
\begin {align*} \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {(7 e) \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{6 b}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (7 b e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{5/2}}{\left (a b+b^2 x\right )^3} \, dx}{6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^2 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{\left (a b+b^2 x\right )^2} \, dx}{24 b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^3 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{16 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^3 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{16 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^2 \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{8 b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^{7/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {35 e^3 (a+b x) \sqrt {d+e x}}{8 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{24 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{12 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^3 \sqrt {b d-a e} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [C] time = 0.04, size = 68, normalized size = 0.28 \begin {gather*} \frac {2 e^3 (a+b x) (d+e x)^{9/2} \, _2F_1\left (4,\frac {9}{2};\frac {11}{2};-\frac {b (d+e x)}{a e-b d}\right )}{9 \sqrt {(a+b x)^2} (a e-b d)^4} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 48.23, size = 257, normalized size = 1.07 \begin {gather*} \frac {(-a e-b e x) \left (\frac {35 \left (b d e^3-a e^4\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 b^{9/2} \sqrt {a e-b d}}-\frac {e^3 \sqrt {d+e x} \left (105 a^3 e^3+280 a^2 b e^2 (d+e x)-315 a^2 b d e^2+315 a b^2 d^2 e+231 a b^2 e (d+e x)^2-560 a b^2 d e (d+e x)-105 b^3 d^3+280 b^3 d^2 (d+e x)+48 b^3 (d+e x)^3-231 b^3 d (d+e x)^2\right )}{24 b^4 (a e+b (d+e x)-b d)^3}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 498, normalized size = 2.07 \begin {gather*} \left [\frac {105 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\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 (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \, {\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \, {\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, -\frac {105 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\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 (48 \, b^{3} e^{3} x^{3} - 8 \, b^{3} d^{3} - 14 \, a b^{2} d^{2} e - 35 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} - 3 \, {\left (29 \, b^{3} d e^{2} - 77 \, a b^{2} e^{3}\right )} x^{2} - 2 \, {\left (19 \, b^{3} d^{2} e + 49 \, a b^{2} d e^{2} - 140 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.35, size = 320, normalized size = 1.33 \begin {gather*} \frac {35 \, {\left (b d e^{3} - a e^{4}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \sqrt {-b^{2} d + a b e} b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac {2 \, \sqrt {x e + d} e^{3}}{b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac {87 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{3} d e^{3} - 136 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{3} + 57 \, \sqrt {x e + d} b^{3} d^{3} e^{3} - 87 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{2} e^{4} + 272 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} d e^{4} - 171 \, \sqrt {x e + d} a b^{2} d^{2} e^{4} - 136 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b e^{5} + 171 \, \sqrt {x e + d} a^{2} b d e^{5} - 57 \, \sqrt {x e + d} a^{3} e^{6}}{24 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 638, normalized size = 2.65 \begin {gather*} \frac {\left (-105 a \,b^{3} e^{4} x^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 b^{4} d \,e^{3} x^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-315 a^{2} b^{2} e^{4} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+315 a \,b^{3} d \,e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-315 a^{3} b \,e^{4} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+315 a^{2} b^{2} d \,e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-105 a^{4} e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+105 a^{3} b d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+48 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{3} e^{3} x^{3}+144 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a \,b^{2} e^{3} x^{2}+144 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} b \,e^{3} x +105 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}-171 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}+171 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e -57 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}+136 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{2}-272 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d e +136 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2}+87 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e -87 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d \right ) \left (b x +a \right )^{2}}{24 \sqrt {\left (a e -b d \right ) b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{\frac {7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{7/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \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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