Optimal. Leaf size=223 \[ \frac {15 e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}-\frac {15 \sqrt {b} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}+\frac {5 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)} \]
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Rubi [A] time = 0.11, antiderivative size = 223, 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 {15 e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}-\frac {15 \sqrt {b} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}}+\frac {5 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)} \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)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^3 (d+e x)^{3/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{2 (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (5 b e \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 (d+e x)^{3/2}} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e}{4 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 e^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}{8 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e}{4 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 e^2 (a+b x)}{4 (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 b e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e}{4 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 e^2 (a+b x)}{4 (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (15 b e \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 )}{4 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 e}{4 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 e^2 (a+b x)}{4 (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 \sqrt {b} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (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 = 65, normalized size = 0.29 \begin {gather*} \frac {2 e^2 (a+b x) \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{\sqrt {(a+b x)^2} \sqrt {d+e x} (b d-a e)^3} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 37.26, size = 195, normalized size = 0.87 \begin {gather*} \frac {(-a e-b e x) \left (-\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}}\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 = 0.44, size = 782, normalized size = 3.51 \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 = 0.34, size = 499, normalized size = 2.24 \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} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, e^{2}}{{\left (b^{3} d^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\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} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\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 = 0.07, size = 285, normalized size = 1.28 \begin {gather*} -\frac {\left (15 \sqrt {e x +d}\, b^{3} e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+30 \sqrt {e x +d}\, a \,b^{2} e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 \sqrt {e x +d}\, a^{2} b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 \sqrt {\left (a e -b d \right ) b}\, b^{2} e^{2} x^{2}+25 \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x +5 \sqrt {\left (a e -b d \right ) b}\, b^{2} d e x +8 \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}+9 \sqrt {\left (a e -b d \right ) b}\, a b d e -2 \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2}\right ) \left (b x +a \right )}{4 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{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}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {3}{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}{{\left (d+e\,x\right )}^{3/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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