Optimal. Leaf size=219 \[ \frac {2 b^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {2 b (a+b x)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}+\frac {2 (a+b x)}{5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac {2 b^{5/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]
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Rubi [A] time = 0.09, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {646, 51, 63, 208} \begin {gather*} \frac {2 b^2 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {2 b (a+b x)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}+\frac {2 (a+b x)}{5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac {2 b^{5/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 208
Rule 646
Rubi steps
\begin {align*} \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{7/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{5 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{5 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{5 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b^2 (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^3 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{5 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b^2 (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 b^3 \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 )}{e (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{5 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b^2 (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{5/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [C] time = 0.02, size = 64, normalized size = 0.29 \begin {gather*} \frac {2 (a+b x) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\frac {b (d+e x)}{b d-a e}\right )}{5 \sqrt {(a+b x)^2} (d+e x)^{5/2} (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 41.76, size = 169, normalized size = 0.77 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {2 \left (3 a^2 e^2-5 a b e (d+e x)-6 a b d e+3 b^2 d^2+15 b^2 (d+e x)^2+5 b^2 d (d+e x)\right )}{15 (d+e x)^{5/2} (b d-a e)^3}-\frac {2 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{(a e-b d)^{7/2}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.42, size = 706, normalized size = 3.22 \begin {gather*} \left [-\frac {15 \, {\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\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} + 23 \, b^{2} d^{2} - 11 \, a b d e + 3 \, a^{2} e^{2} + 5 \, {\left (7 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} - 3 \, a b^{2} d^{5} e + 3 \, a^{2} b d^{4} e^{2} - a^{3} d^{3} e^{3} + {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4}\right )} x\right )}}, -\frac {2 \, {\left (15 \, {\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\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} + 23 \, b^{2} d^{2} - 11 \, a b d e + 3 \, a^{2} e^{2} + 5 \, {\left (7 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (b^{3} d^{6} - 3 \, a b^{2} d^{5} e + 3 \, a^{2} b d^{4} e^{2} - a^{3} d^{3} e^{3} + {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4}\right )} x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.19, size = 196, normalized size = 0.89 \begin {gather*} \frac {2}{15} \, {\left (\frac {15 \, b^{3} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\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 {15 \, {\left (x e + d\right )}^{2} b^{2} + 5 \, {\left (x e + d\right )} b^{2} d + 3 \, b^{2} d^{2} - 5 \, {\left (x e + d\right )} a b e - 6 \, a b d e + 3 \, a^{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 )} {\left (x e + d\right )}^{\frac {5}{2}}}\right )} \mathrm {sgn}\left (b x + a\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 202, normalized size = 0.92 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (15 \sqrt {\left (a e -b d \right ) b}\, b^{2} e^{2} x^{2}-5 \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x +35 \sqrt {\left (a e -b d \right ) b}\, b^{2} d e x +3 \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}-11 \sqrt {\left (a e -b d \right ) b}\, a b d e +15 \left (e x +d \right )^{\frac {5}{2}} b^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+23 \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}}} \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 {1}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {7}{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 {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{7/2}} \,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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