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

Optimal. Leaf size=41 \[ \frac {1}{8} a c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b c^2 x^8 \sqrt {c x^2} \]

[Out]

1/8*a*c^2*x^7*(c*x^2)^(1/2)+1/9*b*c^2*x^8*(c*x^2)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 41, 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} \[ \frac {1}{8} a c^2 x^7 \sqrt {c x^2}+\frac {1}{9} b c^2 x^8 \sqrt {c x^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(c*x^2)^(5/2)*(a + b*x),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.59 \[ \frac {1}{72} x^3 \left (c x^2\right )^{5/2} (9 a+8 b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*(c*x^2)^(5/2)*(9*a + 8*b*x))/72

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fricas [A]  time = 0.45, size = 28, normalized size = 0.68 \[ \frac {1}{72} \, {\left (8 \, b c^{2} x^{8} + 9 \, a c^{2} x^{7}\right )} \sqrt {c x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(8*b*c^2*x^8 + 9*a*c^2*x^7)*sqrt(c*x^2)

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giac [A]  time = 0.80, size = 28, normalized size = 0.68 \[ \frac {1}{72} \, {\left (8 \, b c^{2} x^{9} \mathrm {sgn}\relax (x) + 9 \, a c^{2} x^{8} \mathrm {sgn}\relax (x)\right )} \sqrt {c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(8*b*c^2*x^9*sgn(x) + 9*a*c^2*x^8*sgn(x))*sqrt(c)

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maple [A]  time = 0.00, size = 21, normalized size = 0.51 \[ \frac {\left (8 b x +9 a \right ) \left (c \,x^{2}\right )^{\frac {5}{2}} x^{3}}{72} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*x^3*(8*b*x+9*a)*(c*x^2)^(5/2)

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maxima [A]  time = 1.30, size = 31, normalized size = 0.76 \[ \frac {\left (c x^{2}\right )^{\frac {7}{2}} b x^{2}}{9 \, c} + \frac {\left (c x^{2}\right )^{\frac {7}{2}} a x}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(c*x^2)^(7/2)*b*x^2/c + 1/8*(c*x^2)^(7/2)*a*x/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^2)^(5/2)*(a + b*x),x)

[Out]

int(x^2*(c*x^2)^(5/2)*(a + b*x), x)

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sympy [A]  time = 2.10, size = 36, normalized size = 0.88 \[ \frac {a c^{\frac {5}{2}} x^{3} \left (x^{2}\right )^{\frac {5}{2}}}{8} + \frac {b c^{\frac {5}{2}} x^{4} \left (x^{2}\right )^{\frac {5}{2}}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c**(5/2)*x**3*(x**2)**(5/2)/8 + b*c**(5/2)*x**4*(x**2)**(5/2)/9

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