Optimal. Leaf size=329 \[ \frac {63 b^2 e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^5}+\frac {21 b e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^4}+\frac {63 e^2 (a+b x)}{20 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^3}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}+\frac {9 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}-\frac {63 b^{5/2} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}} \]
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Rubi [A] time = 0.16, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {646, 51, 63, 208} \[ \frac {63 b^2 e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^5}+\frac {21 b e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^4}+\frac {63 e^2 (a+b x)}{20 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^3}-\frac {63 b^{5/2} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}+\frac {9 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2} \]
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} \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)^{7/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (9 b e \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 (d+e x)^{7/2}} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{7/2}} \, dx}{8 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b^2 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)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 b^2 e^2 (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b^3 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)^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 b^2 e^2 (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (63 b^3 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)^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {9 e}{4 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 e^2 (a+b x)}{20 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {21 b e^2 (a+b x)}{4 (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {63 b^2 e^2 (a+b x)}{4 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {63 b^{5/2} 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)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [C] time = 0.03, size = 67, normalized size = 0.20 \[ \frac {2 e^2 (a+b x) \, _2F_1\left (-\frac {5}{2},3;-\frac {3}{2};\frac {b (d+e x)}{b d-a e}\right )}{5 \sqrt {(a+b x)^2} (d+e x)^{5/2} (b d-a e)^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.06, size = 1858, normalized size = 5.65 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.46, size = 775, normalized size = 2.36 \[ \frac {63 \, b^{3} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{5} d^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{5} e^{5} \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 {15 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} e^{2} - 17 \, \sqrt {x e + d} b^{4} d e^{2} + 17 \, \sqrt {x e + d} a b^{3} e^{3}}{4 \, {\left (b^{5} d^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{5} e^{5} \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}} + \frac {2 \, {\left (30 \, {\left (x e + d\right )}^{2} b^{2} e^{2} + 5 \, {\left (x e + d\right )} b^{2} d e^{2} + b^{2} d^{2} e^{2} - 5 \, {\left (x e + d\right )} a b e^{3} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}}{5 \, {\left (b^{5} d^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{5} e^{5} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} {\left (x e + d\right )}^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 518, normalized size = 1.57 \[ -\frac {\left (315 \sqrt {\left (a e -b d \right ) b}\, b^{4} e^{4} x^{4}+525 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} e^{4} x^{3}+735 \sqrt {\left (a e -b d \right ) b}\, b^{4} d \,e^{3} x^{3}+168 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} e^{4} x^{2}+1239 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d \,e^{3} x^{2}+315 \left (e x +d \right )^{\frac {5}{2}} b^{5} e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+483 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{2} e^{2} x^{2}-24 \sqrt {\left (a e -b d \right ) b}\, a^{3} b \,e^{4} x +408 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} d \,e^{3} x +630 \left (e x +d \right )^{\frac {5}{2}} a \,b^{4} e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+831 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d^{2} e^{2} x +45 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{3} e x +8 \sqrt {\left (a e -b d \right ) b}\, a^{4} e^{4}-56 \sqrt {\left (a e -b d \right ) b}\, a^{3} b d \,e^{3}+315 \left (e x +d \right )^{\frac {5}{2}} a^{2} b^{3} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+288 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} d^{2} e^{2}+85 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d^{3} e -10 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{4}\right ) \left (b x +a \right )}{20 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \]
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 \[ \int \frac {1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {7}{2}}}\,{d x} \]
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 \[ \int \frac {1}{{\left (d+e\,x\right )}^{7/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]
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 \[ \int \frac {1}{\left (d + e x\right )^{\frac {7}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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