\(\int \frac {(c x^2)^{5/2}}{x^4 (a+b x)} \, dx\) [873]

Optimal result

Integrand size = 20, antiderivative size = 44 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx=\frac {c^2 \sqrt {c x^2}}{b}-\frac {a c^2 \sqrt {c x^2} \log (a+b x)}{b^2 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, 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} \[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx=\frac {c^2 \sqrt {c x^2}}{b}-\frac {a c^2 \sqrt {c x^2} \log (a+b x)}{b^2 x} \]

[In]

Int[(c*x^2)^(5/2)/(x^4*(a + b*x)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx=c^2 \sqrt {c x^2} \left (\frac {1}{b}-\frac {a \log (a+b x)}{b^2 x}\right ) \]

[In]

Integrate[(c*x^2)^(5/2)/(x^4*(a + b*x)),x]

[Out]

c^2*Sqrt[c*x^2]*(b^(-1) - (a*Log[a + b*x])/(b^2*x))

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66

method	result	size
default	\(-\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \left (a \ln \left (b x +a \right )-b x \right )}{b^{2} x^{5}}\)	\(29\)
risch	\(\frac {c^{2} \sqrt {c \,x^{2}}}{b}-\frac {a \,c^{2} \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{2} x}\)	\(41\)

[In]

int((c*x^2)^(5/2)/x^4/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx=\frac {{\left (b c^{2} x - a c^{2} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{b^{2} x} \]

[In]

integrate((c*x^2)^(5/2)/x^4/(b*x+a),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx=\int \frac {\left (c x^{2}\right )^{\frac {5}{2}}}{x^{4} \left (a + b x\right )}\, dx \]

[In]

integrate((c*x**2)**(5/2)/x**4/(b*x+a),x)

[Out]

Integral((c*x**2)**(5/2)/(x**4*(a + b*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.75 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx=-\frac {\left (-1\right )^{\frac {2 \, c x}{b}} a c^{\frac {5}{2}} \log \left (\frac {2 \, c x}{b}\right )}{b^{2}} - \frac {\left (-1\right )^{\frac {2 \, a c x}{b}} a c^{\frac {5}{2}} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{2}} + \frac {\sqrt {c x^{2}} c^{2}}{b} \]

[In]

integrate((c*x^2)^(5/2)/x^4/(b*x+a),x, algorithm="maxima")

[Out]

-(-1)^(2*c*x/b)*a*c^(5/2)*log(2*c*x/b)/b^2 - (-1)^(2*a*c*x/b)*a*c^(5/2)*log(-2*a*c*x/(b*abs(b*x + a)))/b^2 + s
qrt(c*x^2)*c^2/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx={\left (\frac {c^{2} x \mathrm {sgn}\left (x\right )}{b} - \frac {a c^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {a c^{2} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{2}}\right )} \sqrt {c} \]

[In]

integrate((c*x^2)^(5/2)/x^4/(b*x+a),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c x^2\right )^{5/2}}{x^4 (a+b x)} \, dx=\int \frac {{\left (c\,x^2\right )}^{5/2}}{x^4\,\left (a+b\,x\right )} \,d x \]

[In]

int((c*x^2)^(5/2)/(x^4*(a + b*x)),x)

[Out]

int((c*x^2)^(5/2)/(x^4*(a + b*x)), x)