Optimal. Leaf size=230 \[ \frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) (b d-a e)^{3/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rubi [A]
time = 0.10, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {784, 81, 52, 65,
214} \begin {gather*} \frac {2 (a+b x) (d+e x)^{3/2} (A b-a B)}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (a+b x) (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (a+b x) \sqrt {d+e x} (A b-a B) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 52
Rule 65
Rule 81
Rule 214
Rule 784
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^{3/2}}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {(A+B x) (d+e x)^{3/2}}{a b+b^2 x} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{a b+b^2 x} \, dx}{5 b^2 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (b^2 d-a b e\right ) \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{5 b^4 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (2 \left (b^2 d-a b e\right )^2 \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{5 b^6 e \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (4 \left (b^2 d-a b e\right )^2 \left (\frac {5}{2} A b^2 e-\frac {5}{2} a b B e\right ) \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 )}{5 b^6 e^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {2 (A b-a B) (b d-a e) (a+b x) \sqrt {d+e x}}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 (A b-a B) (a+b x) (d+e x)^{3/2}}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 B (a+b x) (d+e x)^{5/2}}{5 b e \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 (A b-a B) (b d-a e)^{3/2} (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [A]
time = 0.23, size = 148, normalized size = 0.64 \begin {gather*} \frac {2 (a+b x) \left (\frac {\sqrt {b} \sqrt {d+e x} \left (15 a^2 B e^2-5 a b e (4 B d+3 A e+B e x)+b^2 \left (3 B (d+e x)^2+5 A e (4 d+e x)\right )\right )}{e}+15 (A b-a B) (-b d+a e)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{15 b^{7/2} \sqrt {(a+b x)^2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(413\) vs.
\(2(166)=332\).
time = 0.80, size = 414, normalized size = 1.80
method | result | size |
risch | \(-\frac {2 \left (-3 B \,b^{2} e^{2} x^{2}-5 A \,b^{2} e^{2} x +5 B a b \,e^{2} x -6 B \,b^{2} d e x +15 A a b \,e^{2}-20 A \,b^{2} d e -15 a^{2} B \,e^{2}+20 B a b d e -3 b^{2} B \,d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 e \,b^{3} \left (b x +a \right )}+\frac {\left (\frac {2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) A \,a^{2} e^{2}}{b^{2} \sqrt {b \left (a e -b d \right )}}-\frac {4 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) A a d e}{b \sqrt {b \left (a e -b d \right )}}+\frac {2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) A \,d^{2}}{\sqrt {b \left (a e -b d \right )}}-\frac {2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) B \,e^{2} a^{3}}{b^{3} \sqrt {b \left (a e -b d \right )}}+\frac {4 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) B \,a^{2} d e}{b^{2} \sqrt {b \left (a e -b d \right )}}-\frac {2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) B a \,d^{2}}{b \sqrt {b \left (a e -b d \right )}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\) | \(397\) |
default | \(\frac {2 \left (b x +a \right ) \left (3 B \sqrt {b \left (a e -b d \right )}\, \left (e x +d \right )^{\frac {5}{2}} b^{2}+5 A \sqrt {b \left (a e -b d \right )}\, \left (e x +d \right )^{\frac {3}{2}} b^{2} e +15 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b \,e^{3}-30 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{2} d \,e^{2}+15 A \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{3} d^{2} e -5 B \sqrt {b \left (a e -b d \right )}\, \left (e x +d \right )^{\frac {3}{2}} a b e -15 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{3} e^{3}+30 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b d \,e^{2}-15 B \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{2} d^{2} e -15 A \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}\, a b \,e^{2}+15 A \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}\, b^{2} d e +15 B \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}\, a^{2} e^{2}-15 B \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}\, a b d e \right )}{15 \sqrt {\left (b x +a \right )^{2}}\, e \,b^{3} \sqrt {b \left (a e -b d \right )}}\) | \(414\) |
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 = 2.31, size = 372, normalized size = 1.62 \begin {gather*} \left [-\frac {{\left (15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {2 \, b d - 2 \, \sqrt {x e + d} b \sqrt {\frac {b d - a e}{b}} + {\left (b x - a\right )} e}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{2} + {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} e^{2} + 2 \, {\left (3 \, B b^{2} d x - 10 \, {\left (B a b - A b^{2}\right )} d\right )} e\right )} \sqrt {x e + d}\right )} e^{\left (-1\right )}}{15 \, b^{3}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} d e - {\left (B a^{2} - A a b\right )} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {x e + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (3 \, B b^{2} d^{2} + {\left (3 \, B b^{2} x^{2} + 15 \, B a^{2} - 15 \, A a b - 5 \, {\left (B a b - A b^{2}\right )} x\right )} e^{2} + 2 \, {\left (3 \, B b^{2} d x - 10 \, {\left (B a b - A b^{2}\right )} d\right )} e\right )} \sqrt {x e + d}\right )} e^{\left (-1\right )}}{15 \, b^{3}}\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 {\left (A + B x\right ) \left (d + e x\right )^{\frac {3}{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 = 1.04, size = 306, normalized size = 1.33 \begin {gather*} -\frac {2 \, {\left (B a b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, B a^{2} b d e \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + B a^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{3} e^{5} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{4} e^{5} \mathrm {sgn}\left (b x + a\right ) - 15 \, \sqrt {x e + d} B a b^{3} d e^{5} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} A b^{4} d e^{5} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} B a^{2} b^{2} e^{6} \mathrm {sgn}\left (b x + a\right ) - 15 \, \sqrt {x e + d} A a b^{3} e^{6} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{15 \, b^{5}} \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 {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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