Optimal. Leaf size=168 \[ \frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rubi [A]
time = 0.06, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {660, 53, 65,
214} \begin {gather*} \frac {2 b (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}+\frac {2 (a+b x)}{3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}-\frac {2 b^{3/2} (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)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 53
Rule 65
Rule 214
Rule 660
Rubi steps
\begin {align*} \int \frac {1}{(d+e x)^{5/2} \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)^{5/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \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)^{3/2}} \, dx}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \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 ) \sqrt {d+e x}} \, dx}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 b^2 \left (a b+b^2 x\right )\right ) \text {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)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (a+b x)}{3 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{(b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{3/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [A]
time = 0.18, size = 113, normalized size = 0.67 \begin {gather*} \frac {2 (a+b x) \left (\sqrt {-b d+a e} (4 b d-a e+3 b e x)+3 b^{3/2} (d+e x)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{3 (-b d+a e)^{5/2} \sqrt {(a+b x)^2} (d+e x)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.54, size = 130, normalized size = 0.77
method | result | size |
default | \(\frac {2 \left (b x +a \right ) \left (3 b^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) \left (e x +d \right )^{\frac {3}{2}}+3 \sqrt {b \left (a e -b d \right )}\, b e x -\sqrt {b \left (a e -b d \right )}\, a e +4 \sqrt {b \left (a e -b d \right )}\, b d \right )}{3 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{2} \sqrt {b \left (a e -b d \right )}\, \left (e x +d \right )^{\frac {3}{2}}}\) | \(130\) |
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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 3.20, size = 413, normalized size = 2.46 \begin {gather*} \left [\frac {3 \, {\left (b x^{2} e^{2} + 2 \, b d x e + b d^{2}\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {2 \, b d - 2 \, {\left (b d - a e\right )} \sqrt {x e + d} \sqrt {\frac {b}{b d - a e}} + {\left (b x - a\right )} e}{b x + a}\right ) + 2 \, {\left (4 \, b d + {\left (3 \, b x - a\right )} e\right )} \sqrt {x e + d}}{3 \, {\left (b^{2} d^{4} + a^{2} x^{2} e^{4} - 2 \, {\left (a b d x^{2} - a^{2} d x\right )} e^{3} + {\left (b^{2} d^{2} x^{2} - 4 \, a b d^{2} x + a^{2} d^{2}\right )} e^{2} + 2 \, {\left (b^{2} d^{3} x - a b d^{3}\right )} e\right )}}, -\frac {2 \, {\left (3 \, {\left (b x^{2} e^{2} + 2 \, b d x e + b d^{2}\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {x e + d} \sqrt {-\frac {b}{b d - a e}}}{b x e + b d}\right ) - {\left (4 \, b d + {\left (3 \, b x - a\right )} e\right )} \sqrt {x e + d}\right )}}{3 \, {\left (b^{2} d^{4} + a^{2} x^{2} e^{4} - 2 \, {\left (a b d x^{2} - a^{2} d x\right )} e^{3} + {\left (b^{2} d^{2} x^{2} - 4 \, a b d^{2} x + a^{2} d^{2}\right )} e^{2} + 2 \, {\left (b^{2} d^{3} x - a b d^{3}\right )} e\right )}}\right ] \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 {5}{2}} \sqrt {\left (a + b x\right )^{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.95, size = 126, normalized size = 0.75 \begin {gather*} \frac {2}{3} \, {\left (\frac {3 \, b^{2} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (x e + d\right )} b + b d - a e}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} {\left (x e + d\right )}^{\frac {3}{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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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