Optimal. Leaf size=130 \[ -\frac {2 (A b-a B) (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}}+\frac {2 \sqrt {d+e x} (A b-a B) (b d-a e)}{b^3}+\frac {2 (d+e x)^{3/2} (A b-a B)}{3 b^2}+\frac {2 B (d+e x)^{5/2}}{5 b e} \]
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Rubi [A] time = 0.08, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 208} \begin {gather*} \frac {2 (d+e x)^{3/2} (A b-a B)}{3 b^2}+\frac {2 \sqrt {d+e x} (A b-a B) (b d-a e)}{b^3}-\frac {2 (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}}+\frac {2 B (d+e x)^{5/2}}{5 b e} \end {gather*}
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 80
Rule 208
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^{3/2}}{a+b x} \, dx &=\frac {2 B (d+e x)^{5/2}}{5 b e}+\frac {\left (2 \left (\frac {5 A b e}{2}-\frac {5 a B e}{2}\right )\right ) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{5 b e}\\ &=\frac {2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac {2 B (d+e x)^{5/2}}{5 b e}+\frac {((A b-a B) (b d-a e)) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{b^2}\\ &=\frac {2 (A b-a B) (b d-a e) \sqrt {d+e x}}{b^3}+\frac {2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac {2 B (d+e x)^{5/2}}{5 b e}+\frac {\left ((A b-a B) (b d-a e)^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{b^3}\\ &=\frac {2 (A b-a B) (b d-a e) \sqrt {d+e x}}{b^3}+\frac {2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac {2 B (d+e x)^{5/2}}{5 b e}+\frac {\left (2 (A b-a B) (b d-a 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 )}{b^3 e}\\ &=\frac {2 (A b-a B) (b d-a e) \sqrt {d+e x}}{b^3}+\frac {2 (A b-a B) (d+e x)^{3/2}}{3 b^2}+\frac {2 B (d+e x)^{5/2}}{5 b e}-\frac {2 (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}}\\ \end {align*}
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Mathematica [A] time = 0.23, size = 109, normalized size = 0.84 \begin {gather*} \frac {2 (A b-a B) \left (\sqrt {b} \sqrt {d+e x} (-3 a e+4 b d+b e x)-3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )\right )}{3 b^{7/2}}+\frac {2 B (d+e x)^{5/2}}{5 b e} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.20, size = 154, normalized size = 1.18 \begin {gather*} \frac {2 \sqrt {d+e x} \left (15 a^2 B e^2-15 a A b e^2-5 a b B e (d+e x)-15 a b B d e+5 A b^2 e (d+e x)+15 A b^2 d e+3 b^2 B (d+e x)^2\right )}{15 b^3 e}-\frac {2 (A b-a B) (a e-b d)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.59, size = 373, normalized size = 2.87 \begin {gather*} \left [-\frac {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 {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3} e}, \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 {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) + {\left (3 \, B b^{2} e^{2} x^{2} + 3 \, B b^{2} d^{2} - 20 \, {\left (B a b - A b^{2}\right )} d e + 15 \, {\left (B a^{2} - A a b\right )} e^{2} + {\left (6 \, B b^{2} d e - 5 \, {\left (B a b - A b^{2}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3} e}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 1.31, size = 228, normalized size = 1.75 \begin {gather*} -\frac {2 \, {\left (B a b^{2} d^{2} - A b^{3} d^{2} - 2 \, B a^{2} b d e + 2 \, A a b^{2} d e + B a^{3} e^{2} - A a^{2} b e^{2}\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} - 5 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{3} e^{5} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{4} e^{5} - 15 \, \sqrt {x e + d} B a b^{3} d e^{5} + 15 \, \sqrt {x e + d} A b^{4} d e^{5} + 15 \, \sqrt {x e + d} B a^{2} b^{2} e^{6} - 15 \, \sqrt {x e + d} A a b^{3} e^{6}\right )} e^{\left (-5\right )}}{15 \, b^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 370, normalized size = 2.85 \begin {gather*} \frac {2 A \,a^{2} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {4 A a d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}+\frac {2 A \,d^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}-\frac {2 B \,a^{3} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {4 B \,a^{2} d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {2 B a \,d^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b}-\frac {2 \sqrt {e x +d}\, A a e}{b^{2}}+\frac {2 \sqrt {e x +d}\, A d}{b}+\frac {2 \sqrt {e x +d}\, B \,a^{2} e}{b^{3}}-\frac {2 \sqrt {e x +d}\, B a d}{b^{2}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} A}{3 b}-\frac {2 \left (e x +d \right )^{\frac {3}{2}} B a}{3 b^{2}}+\frac {2 \left (e x +d \right )^{\frac {5}{2}} B}{5 b e} \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 = 0.11, size = 236, normalized size = 1.82 \begin {gather*} \left (\frac {2\,A\,e-2\,B\,d}{3\,b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{3\,b^2\,e^2}\right )\,{\left (d+e\,x\right )}^{3/2}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{-B\,a^3\,e^2+2\,B\,a^2\,b\,d\,e+A\,a^2\,b\,e^2-B\,a\,b^2\,d^2-2\,A\,a\,b^2\,d\,e+A\,b^3\,d^2}\right )\,\left (A\,b-B\,a\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{b^{7/2}}+\frac {2\,B\,{\left (d+e\,x\right )}^{5/2}}{5\,b\,e}-\frac {\left (\frac {2\,A\,e-2\,B\,d}{b\,e}-\frac {2\,B\,\left (a\,e^2-b\,d\,e\right )}{b^2\,e^2}\right )\,\left (a\,e^2-b\,d\,e\right )\,\sqrt {d+e\,x}}{b\,e} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 41.17, size = 139, normalized size = 1.07 \begin {gather*} \frac {2 B \left (d + e x\right )^{\frac {5}{2}}}{5 b e} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (2 A b - 2 B a\right )}{3 b^{2}} + \frac {\sqrt {d + e x} \left (- 2 A a b e + 2 A b^{2} d + 2 B a^{2} e - 2 B a b d\right )}{b^{3}} - \frac {2 \left (- A b + B a\right ) \left (a e - b d\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{4} \sqrt {\frac {a e - b d}{b}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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