Optimal. Leaf size=135 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (-2 a B e+A b e+b B d)}{5 b^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)}{4 b^3}+\frac {B e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]
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Rubi [A] time = 0.14, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 77} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (-2 a B e+A b e+b B d)}{5 b^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)}{4 b^3}+\frac {B e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 77
Rule 770
Rubi steps
\begin {align*} \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right )^3 (A+B x) (d+e x) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(A b-a B) (b d-a e) \left (a b+b^2 x\right )^3}{b^2}+\frac {(b B d+A b e-2 a B e) \left (a b+b^2 x\right )^4}{b^3}+\frac {B e \left (a b+b^2 x\right )^5}{b^4}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {(A b-a B) (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^3}+\frac {(b B d+A b e-2 a B e) (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^3}+\frac {B e (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 144, normalized size = 1.07 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (10 a^3 (3 A (2 d+e x)+B x (3 d+2 e x))+15 a^2 b x (A (6 d+4 e x)+B x (4 d+3 e x))+3 a b^2 x^2 (5 A (4 d+3 e x)+3 B x (5 d+4 e x))+b^3 x^3 (3 A (5 d+4 e x)+2 B x (6 d+5 e x))\right )}{60 (a+b x)} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 1.95, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.41, size = 146, normalized size = 1.08 \begin {gather*} \frac {1}{6} \, B b^{3} e x^{6} + A a^{3} d x + \frac {1}{5} \, {\left (B b^{3} d + {\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} x^{5} + \frac {1}{4} \, {\left ({\left (3 \, B a b^{2} + A b^{3}\right )} d + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e\right )} x^{4} + \frac {1}{3} \, {\left (3 \, {\left (B a^{2} b + A a b^{2}\right )} d + {\left (B a^{3} + 3 \, A a^{2} b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a^{3} e + {\left (B a^{3} + 3 \, A a^{2} b\right )} d\right )} x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.17, size = 267, normalized size = 1.98 \begin {gather*} \frac {1}{6} \, B b^{3} x^{6} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B b^{3} d x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, B a b^{2} x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b^{3} x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a b^{2} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a^{2} b x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, A a b^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + B a^{2} b d x^{3} \mathrm {sgn}\left (b x + a\right ) + A a b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a^{3} x^{3} e \mathrm {sgn}\left (b x + a\right ) + A a^{2} b x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a^{3} d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, A a^{2} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{3} x^{2} e \mathrm {sgn}\left (b x + a\right ) + A a^{3} d x \mathrm {sgn}\left (b x + a\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 180, normalized size = 1.33 \begin {gather*} \frac {\left (10 B \,b^{3} e \,x^{5}+12 x^{4} A \,b^{3} e +36 x^{4} B e a \,b^{2}+12 x^{4} B \,b^{3} d +45 x^{3} A a \,b^{2} e +15 x^{3} A d \,b^{3}+45 x^{3} B e \,a^{2} b +45 x^{3} B a \,b^{2} d +60 x^{2} A \,a^{2} b e +60 x^{2} A d a \,b^{2}+20 x^{2} B e \,a^{3}+60 x^{2} B \,a^{2} b d +30 x A \,a^{3} e +90 x A d \,a^{2} b +30 x B \,a^{3} d +60 A d \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{60 \left (b x +a \right )^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.54, size = 254, normalized size = 1.88 \begin {gather*} \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A d x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{2} e x}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a d}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3} e}{4 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (B d + A e\right )} a x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B e x}{6 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (B d + A e\right )} a^{2}}{4 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a e}{30 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (B d + A e\right )}}{5 \, b^{2}} \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 \left (A+B\,x\right )\,\left (d+e\,x\right )\,{\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] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + B x\right ) \left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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