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

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

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 32} \begin {gather*} \frac {c \sqrt {c x^2} (a+b x)^{n+1}}{b (n+1) x} \end {gather*}

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*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.97 \begin {gather*} \frac {\left (c x^2\right )^{3/2} (a+b x)^{n+1}}{b (n+1) x^3} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.21, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^2\right )^{3/2} (a+b x)^n}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.01, size = 33, normalized size = 1.06 \begin {gather*} \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 {gather*}

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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giac [A]  time = 1.09, size = 42, normalized size = 1.35 \begin {gather*} -c^{\frac {3}{2}} {\left (\frac {a^{n + 1} \mathrm {sgn}\relax (x)}{b n + b} - \frac {{\left (b x + a\right )}^{n + 1} \mathrm {sgn}\relax (x)}{b {\left (n + 1\right )}}\right )} \end {gather*}

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)))

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maple [A]  time = 0.00, size = 29, normalized size = 0.94 \begin {gather*} \frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b x +a \right )^{n +1}}{\left (n +1\right ) b \,x^{3}} \end {gather*}

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

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maxima [A]  time = 1.42, size = 28, normalized size = 0.90 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.23, size = 45, normalized size = 1.45 \begin {gather*} \frac {\left (\frac {c\,x\,\sqrt {c\,x^2}}{n+1}+\frac {a\,c\,\sqrt {c\,x^2}}{b\,\left (n+1\right )}\right )\,{\left (a+b\,x\right )}^n}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}}{a x^{2}} & \text {for}\: b = 0 \wedge n = -1 \\\frac {a^{n} c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}}{x^{2}} & \text {for}\: b = 0 \\\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{3} \left (a + b x\right )}\, dx & \text {for}\: n = -1 \\\frac {a c^{\frac {3}{2}} \left (a + b x\right )^{n} \left (x^{2}\right )^{\frac {3}{2}}}{b n x^{3} + b x^{3}} + \frac {b c^{\frac {3}{2}} x \left (a + b x\right )^{n} \left (x^{2}\right )^{\frac {3}{2}}}{b n x^{3} + b x^{3}} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((c**(3/2)*(x**2)**(3/2)/(a*x**2), Eq(b, 0) & Eq(n, -1)), (a**n*c**(3/2)*(x**2)**(3/2)/x**2, Eq(b, 0)
), (Integral((c*x**2)**(3/2)/(x**3*(a + b*x)), x), Eq(n, -1)), (a*c**(3/2)*(a + b*x)**n*(x**2)**(3/2)/(b*n*x**
3 + b*x**3) + b*c**(3/2)*x*(a + b*x)**n*(x**2)**(3/2)/(b*n*x**3 + b*x**3), True))

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