Optimal. Leaf size=280 \[ -\frac {(d+e x)^{5/2} (A b-a B)}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {(d+e x)^{3/2} (-5 a B e+A b e+4 b B d)}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {3 e (a+b x) (-5 a B e+A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}}+\frac {3 e (a+b x) \sqrt {d+e x} (-5 a B e+A b e+4 b B d)}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]
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Rubi [A] time = 0.23, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {770, 78, 47, 50, 63, 208} \begin {gather*} -\frac {(d+e x)^{5/2} (A b-a B)}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {(d+e x)^{3/2} (-5 a B e+A b e+4 b B d)}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac {3 e (a+b x) \sqrt {d+e x} (-5 a B e+A b e+4 b B d)}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {3 e (a+b x) (-5 a B e+A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {b d-a e}} \end {gather*}
Antiderivative was successfully verified.
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Rule 47
Rule 50
Rule 63
Rule 78
Rule 208
Rule 770
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^{3/2}}{\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 {(A+B x) (d+e x)^{3/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((4 b B d+A b e-5 a B e) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^{3/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (3 e (4 b B d+A b e-5 a B e) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{a b+b^2 x} \, dx}{8 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 e (4 b B d+A b e-5 a B e) (a+b x) \sqrt {d+e x}}{4 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (3 e \left (b^2 d-a b e\right ) (4 b B d+A b e-5 a B e) \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^4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 e (4 b B d+A b e-5 a B e) (a+b x) \sqrt {d+e x}}{4 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (3 \left (b^2 d-a b e\right ) (4 b B d+A b e-5 a B 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^4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 e (4 b B d+A b e-5 a B e) (a+b x) \sqrt {d+e x}}{4 b^3 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(4 b B d+A b e-5 a B e) (d+e x)^{3/2}}{4 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^{5/2}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e (4 b B d+A b e-5 a B e) (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {b d-a e} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [C] time = 0.09, size = 110, normalized size = 0.39 \begin {gather*} \frac {(a+b x) (d+e x)^{5/2} \left (\frac {e (a+b x)^2 (-5 a B e+A b e+4 b B d) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+5 a B-5 A b\right )}{10 b \left ((a+b x)^2\right )^{3/2} (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 53.94, size = 236, normalized size = 0.84 \begin {gather*} \frac {(-a e-b e x) \left (\frac {e \sqrt {d+e x} \left (-15 a^2 B e^2+3 a A b e^2-25 a b B e (d+e x)+27 a b B d e+5 A b^2 e (d+e x)-3 A b^2 d e-12 b^2 B d^2-8 b^2 B (d+e x)^2+20 b^2 B d (d+e x)\right )}{4 b^3 (a e+b (d+e x)-b d)^2}+\frac {3 \left (-5 a B e^2+A b e^2+4 b B d e\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{7/2} \sqrt {a e-b d}}\right )}{e \sqrt {\frac {(a e+b e x)^2}{e^2}}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.44, size = 703, normalized size = 2.51 \begin {gather*} \left [\frac {3 \, {\left (4 \, B a^{2} b d e - {\left (5 \, B a^{3} - A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (5 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (5 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d e + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} e^{2} - 8 \, {\left (B b^{4} d e - B a b^{3} e^{2}\right )} x^{2} + {\left (4 \, B b^{4} d^{2} - {\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d e + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{5} d - a^{3} b^{4} e + {\left (b^{7} d - a b^{6} e\right )} x^{2} + 2 \, {\left (a b^{6} d - a^{2} b^{5} e\right )} x\right )}}, \frac {3 \, {\left (4 \, B a^{2} b d e - {\left (5 \, B a^{3} - A a^{2} b\right )} e^{2} + {\left (4 \, B b^{3} d e - {\left (5 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (4 \, B a b^{2} d e - {\left (5 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} - {\left (17 \, B a^{2} b^{2} - A a b^{3}\right )} d e + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} e^{2} - 8 \, {\left (B b^{4} d e - B a b^{3} e^{2}\right )} x^{2} + {\left (4 \, B b^{4} d^{2} - {\left (29 \, B a b^{3} - 5 \, A b^{4}\right )} d e + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{5} d - a^{3} b^{4} e + {\left (b^{7} d - a b^{6} e\right )} x^{2} + 2 \, {\left (a b^{6} d - a^{2} b^{5} e\right )} x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 312, normalized size = 1.11 \begin {gather*} \frac {2 \, \sqrt {x e + d} B e}{b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac {3 \, {\left (4 \, B b d e^{2} - 5 \, B a e^{3} + A b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{4 \, \sqrt {-b^{2} d + a b e} b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac {{\left (4 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} d e^{2} - 4 \, \sqrt {x e + d} B b^{2} d^{2} e^{2} - 9 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{3} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{3} + 11 \, \sqrt {x e + d} B a b d e^{3} - 3 \, \sqrt {x e + d} A b^{2} d e^{3} - 7 \, \sqrt {x e + d} B a^{2} e^{4} + 3 \, \sqrt {x e + d} A a b e^{4}\right )} e^{\left (-1\right )}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 608, normalized size = 2.17 \begin {gather*} -\frac {\left (-3 A \,b^{3} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 B a \,b^{2} e^{3} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-12 B \,b^{3} d \,e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-6 A a \,b^{2} e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+30 B \,a^{2} b \,e^{3} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-24 B a \,b^{2} d \,e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-3 A \,a^{2} b \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 B \,a^{3} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-12 B \,a^{2} b d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-8 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B \,b^{2} e^{2} x^{2}-16 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B a b \,e^{2} x +3 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A a b \,e^{2}-3 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, A \,b^{2} d e -15 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B \,a^{2} e^{2}+11 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B a b d e -4 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, B \,b^{2} d^{2}+5 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} A \,b^{2} e -9 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} B a b e +4 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} B \,b^{2} d \right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{3} e} \end {gather*}
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*} \int \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}\,{d x} \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}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \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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