\(\int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx\) [877]

Optimal result

Integrand size = 20, antiderivative size = 83 \[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=\frac {a^2 x^2}{b^3 \sqrt {c x^2}}-\frac {a x^3}{2 b^2 \sqrt {c x^2}}+\frac {x^4}{3 b \sqrt {c x^2}}-\frac {a^3 x \log (a+b x)}{b^4 \sqrt {c x^2}} \]

[Out]

a^2*x^2/b^3/(c*x^2)^(1/2)-1/2*a*x^3/b^2/(c*x^2)^(1/2)+1/3*x^4/b/(c*x^2)^(1/2)-a^3*x*ln(b*x+a)/b^4/(c*x^2)^(1/2
)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=-\frac {a^3 x \log (a+b x)}{b^4 \sqrt {c x^2}}+\frac {a^2 x^2}{b^3 \sqrt {c x^2}}-\frac {a x^3}{2 b^2 \sqrt {c x^2}}+\frac {x^4}{3 b \sqrt {c x^2}} \]

[In]

Int[x^4/(Sqrt[c*x^2]*(a + b*x)),x]

[Out]

(a^2*x^2)/(b^3*Sqrt[c*x^2]) - (a*x^3)/(2*b^2*Sqrt[c*x^2]) + x^4/(3*b*Sqrt[c*x^2]) - (a^3*x*Log[a + b*x])/(b^4*
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 45

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=\frac {x \left (b x \left (6 a^2-3 a b x+2 b^2 x^2\right )-6 a^3 \log (a+b x)\right )}{6 b^4 \sqrt {c x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.60

method	result	size
default	\(-\frac {x \left (-2 b^{3} x^{3}+3 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )-6 a^{2} b x \right )}{6 \sqrt {c \,x^{2}}\, b^{4}}\)	\(50\)
risch	\(\frac {x \left (\frac {1}{3} b^{2} x^{3}-\frac {1}{2} a b \,x^{2}+a^{2} x \right )}{\sqrt {c \,x^{2}}\, b^{3}}-\frac {a^{3} x \ln \left (b x +a \right )}{b^{4} \sqrt {c \,x^{2}}}\)	\(57\)

[In]

int(x^4/(b*x+a)/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*x*(-2*b^3*x^3+3*a*b^2*x^2+6*a^3*ln(b*x+a)-6*a^2*b*x)/(c*x^2)^(1/2)/b^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.65 \[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=\frac {{\left (2 \, b^{3} x^{3} - 3 \, a b^{2} x^{2} + 6 \, a^{2} b x - 6 \, a^{3} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{6 \, b^{4} c x} \]

[In]

integrate(x^4/(b*x+a)/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*b^3*x^3 - 3*a*b^2*x^2 + 6*a^2*b*x - 6*a^3*log(b*x + a))*sqrt(c*x^2)/(b^4*c*x)

Sympy [F]

\[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=\int \frac {x^{4}}{\sqrt {c x^{2}} \left (a + b x\right )}\, dx \]

[In]

integrate(x**4/(b*x+a)/(c*x**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.71 \[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=\frac {\sqrt {c x^{2}} x^{2}}{3 \, b c} - \frac {7 \, a x^{2}}{6 \, b^{2} \sqrt {c}} - \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} a^{3} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{4} \sqrt {c}} + \frac {2 \, \sqrt {c x^{2}} a x}{3 \, b^{2} c} - \frac {14 \, a^{2} x}{3 \, b^{3} \sqrt {c}} - \frac {a^{3} \log \left (b x\right )}{b^{4} \sqrt {c}} + \frac {17 \, \sqrt {c x^{2}} a^{2}}{3 \, b^{3} c} - \frac {7 \, a^{3}}{2 \, b^{4} \sqrt {c}} \]

[In]

integrate(x^4/(b*x+a)/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(c*x^2)*x^2/(b*c) - 7/6*a*x^2/(b^2*sqrt(c)) - (-1)^(2*a*c*x/b)*a^3*log(-2*a*c*x/(b*abs(b*x + a)))/(b^4
*sqrt(c)) + 2/3*sqrt(c*x^2)*a*x/(b^2*c) - 14/3*a^2*x/(b^3*sqrt(c)) - a^3*log(b*x)/(b^4*sqrt(c)) + 17/3*sqrt(c*
x^2)*a^2/(b^3*c) - 7/2*a^3/(b^4*sqrt(c))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=\frac {a^{3} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{4} \sqrt {c}} - \frac {a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{4} \sqrt {c} \mathrm {sgn}\left (x\right )} + \frac {2 \, b^{2} c x^{3} - 3 \, a b c x^{2} + 6 \, a^{2} c x}{6 \, b^{3} c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^4/(b*x+a)/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

a^3*log(abs(a))*sgn(x)/(b^4*sqrt(c)) - a^3*log(abs(b*x + a))/(b^4*sqrt(c)*sgn(x)) + 1/6*(2*b^2*c*x^3 - 3*a*b*c
*x^2 + 6*a^2*c*x)/(b^3*c^(3/2)*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {c x^2} (a+b x)} \, dx=\int \frac {x^4}{\sqrt {c\,x^2}\,\left (a+b\,x\right )} \,d x \]

[In]

int(x^4/((c*x^2)^(1/2)*(a + b*x)),x)

[Out]

int(x^4/((c*x^2)^(1/2)*(a + b*x)), x)