3.815 \(\int (c x^2)^{3/2} (a+b x)^2 \, dx\)

Optimal. Leaf size=60 \[ \frac{1}{4} a^2 c x^3 \sqrt{c x^2}+\frac{2}{5} a b c x^4 \sqrt{c x^2}+\frac{1}{6} b^2 c x^5 \sqrt{c x^2} \]

[Out]

(a^2*c*x^3*Sqrt[c*x^2])/4 + (2*a*b*c*x^4*Sqrt[c*x^2])/5 + (b^2*c*x^5*Sqrt[c*x^2])/6

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Rubi [A]  time = 0.0146034, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {15, 43} \[ \frac{1}{4} a^2 c x^3 \sqrt{c x^2}+\frac{2}{5} a b c x^4 \sqrt{c x^2}+\frac{1}{6} b^2 c x^5 \sqrt{c x^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(3/2)*(a + b*x)^2,x]

[Out]

(a^2*c*x^3*Sqrt[c*x^2])/4 + (2*a*b*c*x^4*Sqrt[c*x^2])/5 + (b^2*c*x^5*Sqrt[c*x^2])/6

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 \left (c x^2\right )^{3/2} (a+b x)^2 \, dx &=\frac{\left (c \sqrt{c x^2}\right ) \int x^3 (a+b x)^2 \, dx}{x}\\ &=\frac{\left (c \sqrt{c x^2}\right ) \int \left (a^2 x^3+2 a b x^4+b^2 x^5\right ) \, dx}{x}\\ &=\frac{1}{4} a^2 c x^3 \sqrt{c x^2}+\frac{2}{5} a b c x^4 \sqrt{c x^2}+\frac{1}{6} b^2 c x^5 \sqrt{c x^2}\\ \end{align*}

Mathematica [A]  time = 0.0065863, size = 33, normalized size = 0.55 \[ \frac{1}{60} x \left (c x^2\right )^{3/2} \left (15 a^2+24 a b x+10 b^2 x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x^2)^(3/2)*(a + b*x)^2,x]

[Out]

(x*(c*x^2)^(3/2)*(15*a^2 + 24*a*b*x + 10*b^2*x^2))/60

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Maple [A]  time = 0.003, size = 30, normalized size = 0.5 \begin{align*}{\frac{x \left ( 10\,{b}^{2}{x}^{2}+24\,abx+15\,{a}^{2} \right ) }{60} \left ( c{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(3/2)*(b*x+a)^2,x)

[Out]

1/60*x*(10*b^2*x^2+24*a*b*x+15*a^2)*(c*x^2)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)*(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54387, size = 85, normalized size = 1.42 \begin{align*} \frac{1}{60} \,{\left (10 \, b^{2} c x^{5} + 24 \, a b c x^{4} + 15 \, a^{2} c x^{3}\right )} \sqrt{c x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)*(b*x+a)^2,x, algorithm="fricas")

[Out]

1/60*(10*b^2*c*x^5 + 24*a*b*c*x^4 + 15*a^2*c*x^3)*sqrt(c*x^2)

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Sympy [A]  time = 0.73236, size = 60, normalized size = 1. \begin{align*} \frac{a^{2} c^{\frac{3}{2}} x \left (x^{2}\right )^{\frac{3}{2}}}{4} + \frac{2 a b c^{\frac{3}{2}} x^{2} \left (x^{2}\right )^{\frac{3}{2}}}{5} + \frac{b^{2} c^{\frac{3}{2}} x^{3} \left (x^{2}\right )^{\frac{3}{2}}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2)**(3/2)*(b*x+a)**2,x)

[Out]

a**2*c**(3/2)*x*(x**2)**(3/2)/4 + 2*a*b*c**(3/2)*x**2*(x**2)**(3/2)/5 + b**2*c**(3/2)*x**3*(x**2)**(3/2)/6

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Giac [A]  time = 1.0478, size = 47, normalized size = 0.78 \begin{align*} \frac{1}{60} \,{\left (10 \, b^{2} x^{6} \mathrm{sgn}\left (x\right ) + 24 \, a b x^{5} \mathrm{sgn}\left (x\right ) + 15 \, a^{2} x^{4} \mathrm{sgn}\left (x\right )\right )} c^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)*(b*x+a)^2,x, algorithm="giac")

[Out]

1/60*(10*b^2*x^6*sgn(x) + 24*a*b*x^5*sgn(x) + 15*a^2*x^4*sgn(x))*c^(3/2)