3.140 \(\int x^{7/2} (A+B x) (b x+c x^2) \, dx\)

Optimal. Leaf size=39 \[ \frac {2}{13} x^{13/2} (A c+b B)+\frac {2}{11} A b x^{11/2}+\frac {2}{15} B c x^{15/2} \]

[Out]

2/11*A*b*x^(11/2)+2/13*(A*c+B*b)*x^(13/2)+2/15*B*c*x^(15/2)

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {765} \[ \frac {2}{13} x^{13/2} (A c+b B)+\frac {2}{11} A b x^{11/2}+\frac {2}{15} B c x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*(A + B*x)*(b*x + c*x^2),x]

[Out]

(2*A*b*x^(11/2))/11 + (2*(b*B + A*c)*x^(13/2))/13 + (2*B*c*x^(15/2))/15

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int x^{7/2} (A+B x) \left (b x+c x^2\right ) \, dx &=\int \left (A b x^{9/2}+(b B+A c) x^{11/2}+B c x^{13/2}\right ) \, dx\\ &=\frac {2}{11} A b x^{11/2}+\frac {2}{13} (b B+A c) x^{13/2}+\frac {2}{15} B c x^{15/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.85 \[ \frac {2 x^{11/2} (15 A (13 b+11 c x)+11 B x (15 b+13 c x))}{2145} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*(A + B*x)*(b*x + c*x^2),x]

[Out]

(2*x^(11/2)*(15*A*(13*b + 11*c*x) + 11*B*x*(15*b + 13*c*x)))/2145

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fricas [A]  time = 0.76, size = 32, normalized size = 0.82 \[ \frac {2}{2145} \, {\left (143 \, B c x^{7} + 195 \, A b x^{5} + 165 \, {\left (B b + A c\right )} x^{6}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(c*x^2+b*x),x, algorithm="fricas")

[Out]

2/2145*(143*B*c*x^7 + 195*A*b*x^5 + 165*(B*b + A*c)*x^6)*sqrt(x)

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giac [A]  time = 0.22, size = 29, normalized size = 0.74 \[ \frac {2}{15} \, B c x^{\frac {15}{2}} + \frac {2}{13} \, B b x^{\frac {13}{2}} + \frac {2}{13} \, A c x^{\frac {13}{2}} + \frac {2}{11} \, A b x^{\frac {11}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(c*x^2+b*x),x, algorithm="giac")

[Out]

2/15*B*c*x^(15/2) + 2/13*B*b*x^(13/2) + 2/13*A*c*x^(13/2) + 2/11*A*b*x^(11/2)

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maple [A]  time = 0.05, size = 28, normalized size = 0.72 \[ \frac {2 \left (143 B c \,x^{2}+165 A c x +165 B b x +195 A b \right ) x^{\frac {11}{2}}}{2145} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)*(c*x^2+b*x),x)

[Out]

2/2145*x^(11/2)*(143*B*c*x^2+165*A*c*x+165*B*b*x+195*A*b)

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maxima [A]  time = 0.47, size = 27, normalized size = 0.69 \[ \frac {2}{15} \, B c x^{\frac {15}{2}} + \frac {2}{11} \, A b x^{\frac {11}{2}} + \frac {2}{13} \, {\left (B b + A c\right )} x^{\frac {13}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(c*x^2+b*x),x, algorithm="maxima")

[Out]

2/15*B*c*x^(15/2) + 2/11*A*b*x^(11/2) + 2/13*(B*b + A*c)*x^(13/2)

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mupad [B]  time = 0.05, size = 27, normalized size = 0.69 \[ \frac {2\,x^{11/2}\,\left (195\,A\,b+165\,A\,c\,x+165\,B\,b\,x+143\,B\,c\,x^2\right )}{2145} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(b*x + c*x^2)*(A + B*x),x)

[Out]

(2*x^(11/2)*(195*A*b + 165*A*c*x + 165*B*b*x + 143*B*c*x^2))/2145

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sympy [A]  time = 7.89, size = 46, normalized size = 1.18 \[ \frac {2 A b x^{\frac {11}{2}}}{11} + \frac {2 A c x^{\frac {13}{2}}}{13} + \frac {2 B b x^{\frac {13}{2}}}{13} + \frac {2 B c x^{\frac {15}{2}}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*(B*x+A)*(c*x**2+b*x),x)

[Out]

2*A*b*x**(11/2)/11 + 2*A*c*x**(13/2)/13 + 2*B*b*x**(13/2)/13 + 2*B*c*x**(15/2)/15

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