Optimal. Leaf size=200 \[ \frac {105 b^{3/2} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{11/2}}-\frac {105 b e^3}{8 \sqrt {d+e x} (b d-a e)^5}-\frac {35 e^3}{8 (d+e x)^{3/2} (b d-a e)^4}-\frac {21 e^2}{8 (a+b x) (d+e x)^{3/2} (b d-a e)^3}+\frac {3 e}{4 (a+b x)^2 (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)} \]
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Rubi [A] time = 0.14, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 51, 63, 208} \begin {gather*} \frac {105 b^{3/2} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{11/2}}-\frac {105 b e^3}{8 \sqrt {d+e x} (b d-a e)^5}-\frac {35 e^3}{8 (d+e x)^{3/2} (b d-a e)^4}-\frac {21 e^2}{8 (a+b x) (d+e x)^{3/2} (b d-a e)^3}+\frac {3 e}{4 (a+b x)^2 (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (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 {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a+b x)^4 (d+e x)^{5/2}} \, dx\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}-\frac {(3 e) \int \frac {1}{(a+b x)^3 (d+e x)^{5/2}} \, dx}{2 (b d-a e)}\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}+\frac {\left (21 e^2\right ) \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{8 (b d-a e)^2}\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {\left (105 e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{16 (b d-a e)^3}\\ &=-\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {\left (105 b e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{16 (b d-a e)^4}\\ &=-\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {105 b e^3}{8 (b d-a e)^5 \sqrt {d+e x}}-\frac {\left (105 b^2 e^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 (b d-a e)^5}\\ &=-\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {105 b e^3}{8 (b d-a e)^5 \sqrt {d+e x}}-\frac {\left (105 b^2 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{8 (b d-a e)^5}\\ &=-\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {105 b e^3}{8 (b d-a e)^5 \sqrt {d+e x}}+\frac {105 b^{3/2} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{11/2}}\\ \end {align*}
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Mathematica [C] time = 0.02, size = 52, normalized size = 0.26 \begin {gather*} -\frac {2 e^3 \, _2F_1\left (-\frac {3}{2},4;-\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{3 (d+e x)^{3/2} (a e-b d)^4} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.96, size = 304, normalized size = 1.52 \begin {gather*} -\frac {e^3 \left (16 a^4 e^4-144 a^3 b e^3 (d+e x)-64 a^3 b d e^3+96 a^2 b^2 d^2 e^2-693 a^2 b^2 e^2 (d+e x)^2+432 a^2 b^2 d e^2 (d+e x)-64 a b^3 d^3 e-432 a b^3 d^2 e (d+e x)-840 a b^3 e (d+e x)^3+1386 a b^3 d e (d+e x)^2+16 b^4 d^4+144 b^4 d^3 (d+e x)-693 b^4 d^2 (d+e x)^2-315 b^4 (d+e x)^4+840 b^4 d (d+e x)^3\right )}{24 (d+e x)^{3/2} (b d-a e)^5 (-a e-b (d+e x)+b d)^3}-\frac {105 b^{3/2} e^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 (a e-b d)^{11/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.45, size = 1840, normalized size = 9.20
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.23, size = 427, normalized size = 2.14 \begin {gather*} -\frac {105 \, b^{2} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \sqrt {-b^{2} d + a b e}} - \frac {315 \, {\left (x e + d\right )}^{4} b^{4} e^{3} - 840 \, {\left (x e + d\right )}^{3} b^{4} d e^{3} + 693 \, {\left (x e + d\right )}^{2} b^{4} d^{2} e^{3} - 144 \, {\left (x e + d\right )} b^{4} d^{3} e^{3} - 16 \, b^{4} d^{4} e^{3} + 840 \, {\left (x e + d\right )}^{3} a b^{3} e^{4} - 1386 \, {\left (x e + d\right )}^{2} a b^{3} d e^{4} + 432 \, {\left (x e + d\right )} a b^{3} d^{2} e^{4} + 64 \, a b^{3} d^{3} e^{4} + 693 \, {\left (x e + d\right )}^{2} a^{2} b^{2} e^{5} - 432 \, {\left (x e + d\right )} a^{2} b^{2} d e^{5} - 96 \, a^{2} b^{2} d^{2} e^{5} + 144 \, {\left (x e + d\right )} a^{3} b e^{6} + 64 \, a^{3} b d e^{6} - 16 \, a^{4} e^{7}}{24 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left ({\left (x e + d\right )}^{\frac {3}{2}} b - \sqrt {x e + d} b d + \sqrt {x e + d} a e\right )}^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 319, normalized size = 1.60 \begin {gather*} \frac {55 \sqrt {e x +d}\, a^{2} b^{2} e^{5}}{8 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{3}}-\frac {55 \sqrt {e x +d}\, a \,b^{3} d \,e^{4}}{4 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{3}}+\frac {55 \sqrt {e x +d}\, b^{4} d^{2} e^{3}}{8 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{3}}+\frac {35 \left (e x +d \right )^{\frac {3}{2}} a \,b^{3} e^{4}}{3 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{3}}-\frac {35 \left (e x +d \right )^{\frac {3}{2}} b^{4} d \,e^{3}}{3 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{3}}+\frac {41 \left (e x +d \right )^{\frac {5}{2}} b^{4} e^{3}}{8 \left (a e -b d \right )^{5} \left (b e x +a e \right )^{3}}+\frac {105 b^{2} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right )^{5} \sqrt {\left (a e -b d \right ) b}}+\frac {8 b \,e^{3}}{\left (a e -b d \right )^{5} \sqrt {e x +d}}-\frac {2 e^{3}}{3 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}} \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 = 0.87, size = 334, normalized size = 1.67 \begin {gather*} \frac {\frac {231\,b^2\,e^3\,{\left (d+e\,x\right )}^2}{8\,{\left (a\,e-b\,d\right )}^3}-\frac {2\,e^3}{3\,\left (a\,e-b\,d\right )}+\frac {35\,b^3\,e^3\,{\left (d+e\,x\right )}^3}{{\left (a\,e-b\,d\right )}^4}+\frac {105\,b^4\,e^3\,{\left (d+e\,x\right )}^4}{8\,{\left (a\,e-b\,d\right )}^5}+\frac {6\,b\,e^3\,\left (d+e\,x\right )}{{\left (a\,e-b\,d\right )}^2}}{{\left (d+e\,x\right )}^{3/2}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )+b^3\,{\left (d+e\,x\right )}^{9/2}-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{7/2}+{\left (d+e\,x\right )}^{5/2}\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )}+\frac {105\,b^{3/2}\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^5\,e^5-5\,a^4\,b\,d\,e^4+10\,a^3\,b^2\,d^2\,e^3-10\,a^2\,b^3\,d^3\,e^2+5\,a\,b^4\,d^4\,e-b^5\,d^5\right )}{{\left (a\,e-b\,d\right )}^{11/2}}\right )}{8\,{\left (a\,e-b\,d\right )}^{11/2}} \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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