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

Optimal. Leaf size=31 \[ \frac{c \sqrt{c x^2} (a+b x)^{n+1}}{b (n+1) x} \]

[Out]

(c*Sqrt[c*x^2]*(a + b*x)^(1 + n))/(b*(1 + n)*x)

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Rubi [A]  time = 0.0059132, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 32} \[ \frac{c \sqrt{c x^2} (a+b x)^{n+1}}{b (n+1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c*x^2]*(a + b*x)^(1 + n))/(b*(1 + n)*x)

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0138324, size = 30, normalized size = 0.97 \[ \frac{\left (c x^2\right )^{3/2} (a+b x)^{n+1}}{b (n+1) x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c*x^2)^(3/2)*(a + b*x)^(1 + n))/(b*(1 + n)*x^3)

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Maple [A]  time = 0.001, size = 29, normalized size = 0.9 \begin{align*}{\frac{ \left ( bx+a \right ) ^{1+n}}{b \left ( 1+n \right ){x}^{3}} \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)^n/x^3,x)

[Out]

(b*x+a)^(1+n)/b/(1+n)*(c*x^2)^(3/2)/x^3

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Maxima [A]  time = 0.991608, size = 38, normalized size = 1.23 \begin{align*} \frac{{\left (b c^{\frac{3}{2}} x + a c^{\frac{3}{2}}\right )}{\left (b x + a\right )}^{n}}{b{\left (n + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c^(3/2)*x + a*c^(3/2))*(b*x + a)^n/(b*(n + 1))

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Fricas [A]  time = 1.61113, size = 72, normalized size = 2.32 \begin{align*} \frac{{\left (b c x + a c\right )} \sqrt{c x^{2}}{\left (b x + a\right )}^{n}}{{\left (b n + b\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c*x + a*c)*sqrt(c*x^2)*(b*x + a)^n/((b*n + b)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.05204, size = 57, normalized size = 1.84 \begin{align*} -c^{\frac{3}{2}}{\left (\frac{a^{n + 1} \mathrm{sgn}\left (x\right )}{b n + b} - \frac{{\left (b x + a\right )}^{n + 1} \mathrm{sgn}\left (x\right )}{b{\left (n + 1\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^(3/2)*(a^(n + 1)*sgn(x)/(b*n + b) - (b*x + a)^(n + 1)*sgn(x)/(b*(n + 1)))