Optimal. Leaf size=281 \[ -\frac {A b-a B}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}+\frac {-a B e+5 A b e-4 b B d}{4 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {3 e (a+b x) (a B e-5 A b e+4 b B d)}{4 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {3 e (a+b x) (a B e-5 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]
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Rubi [A] time = 0.26, antiderivative size = 280, 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 = {770, 78, 51, 63, 208} \begin {gather*} -\frac {A b-a B}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}-\frac {a B e-5 A b e+4 b B d}{4 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {3 e (a+b x) (a B e-5 A b e+4 b B d)}{4 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {3 e (a+b x) (a B e-5 A b e+4 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 51
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}{\left (a b+b^2 x\right )^3 (d+e x)^{3/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {A b-a B}{2 b (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((4 b B d-5 A b e+a B e) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 (d+e x)^{3/2}} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (3 e (4 b B d-5 A b e+a B e) \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 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e (4 b B d-5 A b e+a B e) (a+b x)}{4 b (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (3 e (4 b B d-5 A b e+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 d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e (4 b B d-5 A b e+a B e) (a+b x)}{4 b (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (3 (4 b B d-5 A b e+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 d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 b (b d-a e) (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e (4 b B d-5 A b e+a B e) (a+b x)}{4 b (b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (4 b B d-5 A b e+a B e) (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {b} (b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [C] time = 0.07, size = 110, normalized size = 0.39 \begin {gather*} \frac {(a+b x) \left (\frac {e (a+b x)^2 (-a B e+5 A b e-4 b B d) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+a B-A b\right )}{2 b \left ((a+b x)^2\right )^{3/2} \sqrt {d+e x} (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 52.41, size = 328, normalized size = 1.17 \begin {gather*} \frac {(-a e-b e x) \left (-\frac {e \left (8 a^2 A e^3-5 a^2 B e^2 (d+e x)-8 a^2 B d e^2+25 a A b e^2 (d+e x)-16 a A b d e^2+16 a b B d^2 e-15 a b B d e (d+e x)-3 a b B e (d+e x)^2+8 A b^2 d^2 e-25 A b^2 d e (d+e x)+15 A b^2 e (d+e x)^2-8 b^2 B d^3+20 b^2 B d^2 (d+e x)-12 b^2 B d (d+e x)^2\right )}{4 \sqrt {d+e x} (b d-a e)^3 (-a e-b (d+e x)+b d)^2}-\frac {3 \left (a B e^2-5 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 \sqrt {b} (b d-a e)^3 \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 [B] time = 0.55, size = 1410, normalized size = 5.02
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.40, size = 616, normalized size = 2.19 \begin {gather*} -\frac {3 \, {\left (4 \, B b d e^{2} + B a e^{3} - 5 \, A b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{3} d^{3} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \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 {2 \, {\left (B d e^{2} - A e^{3}\right )}}{{\left (b^{3} d^{3} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt {x e + d}} - \frac {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} + 3 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{3} - 7 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{3} - \sqrt {x e + d} B a b d e^{3} + 9 \, \sqrt {x e + d} A b^{2} d e^{3} + 5 \, \sqrt {x e + d} B a^{2} e^{4} - 9 \, \sqrt {x e + d} A a b e^{4}}{4 \, {\left (b^{3} d^{3} e \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm {sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.08, size = 681, normalized size = 2.42 \begin {gather*} -\frac {\left (15 \sqrt {e x +d}\, A \,b^{3} e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-3 \sqrt {e x +d}\, B a \,b^{2} e^{2} x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-12 \sqrt {e x +d}\, B \,b^{3} d e \,x^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+30 \sqrt {e x +d}\, A a \,b^{2} e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-6 \sqrt {e x +d}\, B \,a^{2} b \,e^{2} x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-24 \sqrt {e x +d}\, B a \,b^{2} d e x \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 \sqrt {e x +d}\, A \,a^{2} b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )+15 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} e^{2} x^{2}-3 \sqrt {e x +d}\, B \,a^{3} e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-12 \sqrt {e x +d}\, B \,a^{2} b d e \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )-3 \sqrt {\left (a e -b d \right ) b}\, B a b \,e^{2} x^{2}-12 \sqrt {\left (a e -b d \right ) b}\, B \,b^{2} d e \,x^{2}+25 \sqrt {\left (a e -b d \right ) b}\, A a b \,e^{2} x +5 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} d e x -5 \sqrt {\left (a e -b d \right ) b}\, B \,a^{2} e^{2} x -21 \sqrt {\left (a e -b d \right ) b}\, B a b d e x -4 \sqrt {\left (a e -b d \right ) b}\, B \,b^{2} d^{2} x +8 \sqrt {\left (a e -b d \right ) b}\, A \,a^{2} e^{2}+9 \sqrt {\left (a e -b d \right ) b}\, A a b d e -2 \sqrt {\left (a e -b d \right ) b}\, A \,b^{2} d^{2}-13 \sqrt {\left (a e -b d \right ) b}\, B \,a^{2} d e -2 \sqrt {\left (a e -b d \right ) b}\, B a b \,d^{2}\right ) \left (b x +a \right )}{4 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \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 {B x + A}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\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 {A+B\,x}{{\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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