\(\int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx\) [852]

Optimal result

Integrand size = 20, antiderivative size = 102 \[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=-\frac {a^3 \sqrt {c x^2}}{b^4}+\frac {a^2 x \sqrt {c x^2}}{2 b^3}-\frac {a x^2 \sqrt {c x^2}}{3 b^2}+\frac {x^3 \sqrt {c x^2}}{4 b}+\frac {a^4 \sqrt {c x^2} \log (a+b x)}{b^5 x} \]

[Out]

-a^3*(c*x^2)^(1/2)/b^4+1/2*a^2*x*(c*x^2)^(1/2)/b^3-1/3*a*x^2*(c*x^2)^(1/2)/b^2+1/4*x^3*(c*x^2)^(1/2)/b+a^4*ln(
b*x+a)*(c*x^2)^(1/2)/b^5/x

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 102, 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} \[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=\frac {a^4 \sqrt {c x^2} \log (a+b x)}{b^5 x}-\frac {a^3 \sqrt {c x^2}}{b^4}+\frac {a^2 x \sqrt {c x^2}}{2 b^3}-\frac {a x^2 \sqrt {c x^2}}{3 b^2}+\frac {x^3 \sqrt {c x^2}}{4 b} \]

[In]

Int[(x^3*Sqrt[c*x^2])/(a + b*x),x]

[Out]

-((a^3*Sqrt[c*x^2])/b^4) + (a^2*x*Sqrt[c*x^2])/(2*b^3) - (a*x^2*Sqrt[c*x^2])/(3*b^2) + (x^3*Sqrt[c*x^2])/(4*b)
+ (a^4*Sqrt[c*x^2]*Log[a + b*x])/(b^5*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 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 {\sqrt {c x^2} \int \frac {x^4}{a+b x} \, dx}{x} \\ & = \frac {\sqrt {c x^2} \int \left (-\frac {a^3}{b^4}+\frac {a^2 x}{b^3}-\frac {a x^2}{b^2}+\frac {x^3}{b}+\frac {a^4}{b^4 (a+b x)}\right ) \, dx}{x} \\ & = -\frac {a^3 \sqrt {c x^2}}{b^4}+\frac {a^2 x \sqrt {c x^2}}{2 b^3}-\frac {a x^2 \sqrt {c x^2}}{3 b^2}+\frac {x^3 \sqrt {c x^2}}{4 b}+\frac {a^4 \sqrt {c x^2} \log (a+b x)}{b^5 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.63 \[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=\sqrt {c x^2} \left (\frac {-12 a^3+6 a^2 b x-4 a b^2 x^2+3 b^3 x^3}{12 b^4}+\frac {a^4 \log (a+b x)}{b^5 x}\right ) \]

[In]

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

[Out]

Sqrt[c*x^2]*((-12*a^3 + 6*a^2*b*x - 4*a*b^2*x^2 + 3*b^3*x^3)/(12*b^4) + (a^4*Log[a + b*x])/(b^5*x))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {\sqrt {c \,x^{2}}\, \left (3 b^{4} x^{4}-4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+12 a^{4} \ln \left (b x +a \right )-12 a^{3} b x \right )}{12 b^{5} x}\)	\(63\)
risch	\(\frac {\sqrt {c \,x^{2}}\, \left (\frac {1}{4} b^{3} x^{4}-\frac {1}{3} a \,b^{2} x^{3}+\frac {1}{2} a^{2} b \,x^{2}-a^{3} x \right )}{x \,b^{4}}+\frac {a^{4} \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{5} x}\)	\(72\)

[In]

int(x^3*(c*x^2)^(1/2)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/12*(c*x^2)^(1/2)*(3*b^4*x^4-4*a*b^3*x^3+6*a^2*b^2*x^2+12*a^4*ln(b*x+a)-12*a^3*b*x)/b^5/x

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.61 \[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=\frac {{\left (3 \, b^{4} x^{4} - 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} - 12 \, a^{3} b x + 12 \, a^{4} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{12 \, b^{5} x} \]

[In]

integrate(x^3*(c*x^2)^(1/2)/(b*x+a),x, algorithm="fricas")

[Out]

1/12*(3*b^4*x^4 - 4*a*b^3*x^3 + 6*a^2*b^2*x^2 - 12*a^3*b*x + 12*a^4*log(b*x + a))*sqrt(c*x^2)/(b^5*x)

Sympy [F]

\[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=\int \frac {x^{3} \sqrt {c x^{2}}}{a + b x}\, dx \]

[In]

integrate(x**3*(c*x**2)**(1/2)/(b*x+a),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.25 \[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=\frac {\left (-1\right )^{\frac {2 \, c x}{b}} a^{4} \sqrt {c} \log \left (\frac {2 \, c x}{b}\right )}{b^{5}} + \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} a^{4} \sqrt {c} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{5}} + \frac {\sqrt {c x^{2}} a^{2} x}{2 \, b^{3}} + \frac {\left (c x^{2}\right )^{\frac {3}{2}} x}{4 \, b c} - \frac {\sqrt {c x^{2}} a^{3}}{b^{4}} - \frac {\left (c x^{2}\right )^{\frac {3}{2}} a}{3 \, b^{2} c} \]

[In]

integrate(x^3*(c*x^2)^(1/2)/(b*x+a),x, algorithm="maxima")

[Out]

(-1)^(2*c*x/b)*a^4*sqrt(c)*log(2*c*x/b)/b^5 + (-1)^(2*a*c*x/b)*a^4*sqrt(c)*log(-2*a*c*x/(b*abs(b*x + a)))/b^5
+ 1/2*sqrt(c*x^2)*a^2*x/b^3 + 1/4*(c*x^2)^(3/2)*x/(b*c) - sqrt(c*x^2)*a^3/b^4 - 1/3*(c*x^2)^(3/2)*a/(b^2*c)

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=\frac {1}{12} \, \sqrt {c} {\left (\frac {12 \, a^{4} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{5}} - \frac {12 \, a^{4} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{5}} + \frac {3 \, b^{3} x^{4} \mathrm {sgn}\left (x\right ) - 4 \, a b^{2} x^{3} \mathrm {sgn}\left (x\right ) + 6 \, a^{2} b x^{2} \mathrm {sgn}\left (x\right ) - 12 \, a^{3} x \mathrm {sgn}\left (x\right )}{b^{4}}\right )} \]

[In]

integrate(x^3*(c*x^2)^(1/2)/(b*x+a),x, algorithm="giac")

[Out]

1/12*sqrt(c)*(12*a^4*log(abs(b*x + a))*sgn(x)/b^5 - 12*a^4*log(abs(a))*sgn(x)/b^5 + (3*b^3*x^4*sgn(x) - 4*a*b^
2*x^3*sgn(x) + 6*a^2*b*x^2*sgn(x) - 12*a^3*x*sgn(x))/b^4)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \sqrt {c x^2}}{a+b x} \, dx=\int \frac {x^3\,\sqrt {c\,x^2}}{a+b\,x} \,d x \]

[In]

int((x^3*(c*x^2)^(1/2))/(a + b*x),x)

[Out]

int((x^3*(c*x^2)^(1/2))/(a + b*x), x)