\(\int \frac {(c x^2)^{3/2}}{(a+b x)^2} \, dx\) [902]

Optimal result

Integrand size = 17, antiderivative size = 89 \[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=-\frac {2 a c \sqrt {c x^2}}{b^3}+\frac {c x \sqrt {c x^2}}{2 b^2}+\frac {a^3 c \sqrt {c x^2}}{b^4 x (a+b x)}+\frac {3 a^2 c \sqrt {c x^2} \log (a+b x)}{b^4 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 45} \[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=\frac {a^3 c \sqrt {c x^2}}{b^4 x (a+b x)}+\frac {3 a^2 c \sqrt {c x^2} \log (a+b x)}{b^4 x}-\frac {2 a c \sqrt {c x^2}}{b^3}+\frac {c x \sqrt {c x^2}}{2 b^2} \]

[In]

Int[(c*x^2)^(3/2)/(a + b*x)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=\frac {\left (c x^2\right )^{3/2} \left (2 a^3-4 a^2 b x-3 a b^2 x^2+b^3 x^3+6 a^2 (a+b x) \log (a+b x)\right )}{2 b^4 x^3 (a+b x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (b^{3} x^{3}+6 \ln \left (b x +a \right ) a^{2} b x -3 a \,b^{2} x^{2}+6 a^{3} \ln \left (b x +a \right )-4 a^{2} b x +2 a^{3}\right )}{2 x^{3} b^{4} \left (b x +a \right )}\)	\(76\)
risch	\(\frac {c \sqrt {c \,x^{2}}\, \left (\frac {1}{2} b \,x^{2}-2 a x \right )}{x \,b^{3}}+\frac {a^{3} c \sqrt {c \,x^{2}}}{b^{4} x \left (b x +a \right )}+\frac {3 a^{2} c \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{4} x}\)	\(78\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=\frac {{\left (b^{3} c x^{3} - 3 \, a b^{2} c x^{2} - 4 \, a^{2} b c x + 2 \, a^{3} c + 6 \, {\left (a^{2} b c x + a^{3} c\right )} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{2 \, {\left (b^{5} x^{2} + a b^{4} x\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{\left (a + b x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.29 \[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=\frac {3 \, \left (-1\right )^{\frac {2 \, c x}{b}} a^{2} c^{\frac {3}{2}} \log \left (\frac {2 \, c x}{b}\right )}{b^{4}} + \frac {3 \, \left (-1\right )^{\frac {2 \, a c x}{b}} a^{2} c^{\frac {3}{2}} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{4}} - \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{b^{2} x + a b} + \frac {3 \, \sqrt {c x^{2}} c x}{2 \, b^{2}} - \frac {3 \, \sqrt {c x^{2}} a c}{b^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=\frac {1}{2} \, c^{\frac {3}{2}} {\left (\frac {6 \, a^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {2 \, a^{3} \mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b^{4}} - \frac {2 \, {\left (3 \, a^{2} \log \left ({\left | a \right |}\right ) + a^{2}\right )} \mathrm {sgn}\left (x\right )}{b^{4}} + \frac {b^{2} x^{2} \mathrm {sgn}\left (x\right ) - 4 \, a b x \mathrm {sgn}\left (x\right )}{b^{4}}\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c x^2\right )^{3/2}}{(a+b x)^2} \, dx=\int \frac {{\left (c\,x^2\right )}^{3/2}}{{\left (a+b\,x\right )}^2} \,d x \]

[In]

int((c*x^2)^(3/2)/(a + b*x)^2,x)

[Out]

int((c*x^2)^(3/2)/(a + b*x)^2, x)