3.918 \(\int \frac{x^4}{(c x^2)^{3/2} (a+b x)^2} \, dx\)

Optimal. Leaf size=49 \[ \frac{a x}{b^2 c \sqrt{c x^2} (a+b x)}+\frac{x \log (a+b x)}{b^2 c \sqrt{c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0138987, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {15, 43} \[ \frac{a x}{b^2 c \sqrt{c x^2} (a+b x)}+\frac{x \log (a+b x)}{b^2 c \sqrt{c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

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

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

Mathematica [A]  time = 0.0110324, size = 37, normalized size = 0.76 \[ \frac{x^3 ((a+b x) \log (a+b x)+a)}{b^2 \left (c x^2\right )^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

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

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Maple [A]  time = 0.003, size = 41, normalized size = 0.8 \begin{align*}{\frac{{x}^{3} \left ( b\ln \left ( bx+a \right ) x+a\ln \left ( bx+a \right ) +a \right ) }{{b}^{2} \left ( bx+a \right ) } \left ( c{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(c*x^2)^(3/2)/(b*x+a)^2,x)

[Out]

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

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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(x^4/(c*x^2)^(3/2)/(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38408, size = 95, normalized size = 1.94 \begin{align*} \frac{\sqrt{c x^{2}}{\left ({\left (b x + a\right )} \log \left (b x + a\right ) + a\right )}}{b^{3} c^{2} x^{2} + a b^{2} c^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2)^(3/2)/(b*x+a)^2,x, algorithm="fricas")

[Out]

sqrt(c*x^2)*((b*x + a)*log(b*x + a) + a)/(b^3*c^2*x^2 + a*b^2*c^2*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (c x^{2}\right )^{\frac{3}{2}} \left (a + b x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(c*x**2)**(3/2)/(b*x+a)**2,x)

[Out]

Integral(x**4/((c*x**2)**(3/2)*(a + b*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (c x^{2}\right )^{\frac{3}{2}}{\left (b x + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(c*x^2)^(3/2)/(b*x+a)^2,x, algorithm="giac")

[Out]

integrate(x^4/((c*x^2)^(3/2)*(b*x + a)^2), x)