3.8.78 \(\int x (c x^2)^{5/2} (a+b x)^2 \, dx\)

Optimal. Leaf size=66 \[ \frac {1}{7} a^2 c^2 x^6 \sqrt {c x^2}+\frac {1}{4} a b c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b^2 c^2 x^8 \sqrt {c x^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 43} \begin {gather*} \frac {1}{7} a^2 c^2 x^6 \sqrt {c x^2}+\frac {1}{4} a b c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b^2 c^2 x^8 \sqrt {c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(c*x^2)^(5/2)*(a + b*x)^2,x]

[Out]

(a^2*c^2*x^6*Sqrt[c*x^2])/7 + (a*b*c^2*x^7*Sqrt[c*x^2])/4 + (b^2*c^2*x^8*Sqrt[c*x^2])/9

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int x \left (c x^2\right )^{5/2} (a+b x)^2 \, dx &=\frac {\left (c^2 \sqrt {c x^2}\right ) \int x^6 (a+b x)^2 \, dx}{x}\\ &=\frac {\left (c^2 \sqrt {c x^2}\right ) \int \left (a^2 x^6+2 a b x^7+b^2 x^8\right ) \, dx}{x}\\ &=\frac {1}{7} a^2 c^2 x^6 \sqrt {c x^2}+\frac {1}{4} a b c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b^2 c^2 x^8 \sqrt {c x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.53 \begin {gather*} \frac {1}{252} x^2 \left (c x^2\right )^{5/2} \left (36 a^2+63 a b x+28 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(c*x^2)^(5/2)*(a + b*x)^2,x]

[Out]

(x^2*(c*x^2)^(5/2)*(36*a^2 + 63*a*b*x + 28*b^2*x^2))/252

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IntegrateAlgebraic [A]  time = 0.03, size = 35, normalized size = 0.53 \begin {gather*} \frac {1}{252} x^2 \left (c x^2\right )^{5/2} \left (36 a^2+63 a b x+28 b^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*(c*x^2)^(5/2)*(a + b*x)^2,x]

[Out]

(x^2*(c*x^2)^(5/2)*(36*a^2 + 63*a*b*x + 28*b^2*x^2))/252

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fricas [A]  time = 0.94, size = 42, normalized size = 0.64 \begin {gather*} \frac {1}{252} \, {\left (28 \, b^{2} c^{2} x^{8} + 63 \, a b c^{2} x^{7} + 36 \, a^{2} c^{2} x^{6}\right )} \sqrt {c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^2)^(5/2)*(b*x+a)^2,x, algorithm="fricas")

[Out]

1/252*(28*b^2*c^2*x^8 + 63*a*b*c^2*x^7 + 36*a^2*c^2*x^6)*sqrt(c*x^2)

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giac [A]  time = 1.22, size = 44, normalized size = 0.67 \begin {gather*} \frac {1}{252} \, {\left (28 \, b^{2} c^{2} x^{9} \mathrm {sgn}\relax (x) + 63 \, a b c^{2} x^{8} \mathrm {sgn}\relax (x) + 36 \, a^{2} c^{2} x^{7} \mathrm {sgn}\relax (x)\right )} \sqrt {c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^2)^(5/2)*(b*x+a)^2,x, algorithm="giac")

[Out]

1/252*(28*b^2*c^2*x^9*sgn(x) + 63*a*b*c^2*x^8*sgn(x) + 36*a^2*c^2*x^7*sgn(x))*sqrt(c)

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maple [A]  time = 0.00, size = 32, normalized size = 0.48 \begin {gather*} \frac {\left (28 b^{2} x^{2}+63 a b x +36 a^{2}\right ) \left (c \,x^{2}\right )^{\frac {5}{2}} x^{2}}{252} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*x^2)^(5/2)*(b*x+a)^2,x)

[Out]

1/252*x^2*(28*b^2*x^2+63*a*b*x+36*a^2)*(c*x^2)^(5/2)

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maxima [A]  time = 1.34, size = 49, normalized size = 0.74 \begin {gather*} \frac {\left (c x^{2}\right )^{\frac {7}{2}} b^{2} x^{2}}{9 \, c} + \frac {\left (c x^{2}\right )^{\frac {7}{2}} a b x}{4 \, c} + \frac {\left (c x^{2}\right )^{\frac {7}{2}} a^{2}}{7 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^2)^(5/2)*(b*x+a)^2,x, algorithm="maxima")

[Out]

1/9*(c*x^2)^(7/2)*b^2*x^2/c + 1/4*(c*x^2)^(7/2)*a*b*x/c + 1/7*(c*x^2)^(7/2)*a^2/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x\,{\left (c\,x^2\right )}^{5/2}\,{\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*x^2)^(5/2)*(a + b*x)^2,x)

[Out]

int(x*(c*x^2)^(5/2)*(a + b*x)^2, x)

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sympy [A]  time = 2.19, size = 60, normalized size = 0.91 \begin {gather*} \frac {a^{2} c^{\frac {5}{2}} x^{2} \left (x^{2}\right )^{\frac {5}{2}}}{7} + \frac {a b c^{\frac {5}{2}} x^{3} \left (x^{2}\right )^{\frac {5}{2}}}{4} + \frac {b^{2} c^{\frac {5}{2}} x^{4} \left (x^{2}\right )^{\frac {5}{2}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x**2)**(5/2)*(b*x+a)**2,x)

[Out]

a**2*c**(5/2)*x**2*(x**2)**(5/2)/7 + a*b*c**(5/2)*x**3*(x**2)**(5/2)/4 + b**2*c**(5/2)*x**4*(x**2)**(5/2)/9

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