\(\int \frac {(c x^2)^{5/2}}{x^5 (a+b x)} \, dx\) [874]

Optimal result

Integrand size = 20, antiderivative size = 25 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^5 (a+b x)} \, dx=\frac {c^2 \sqrt {c x^2} \log (a+b x)}{b x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 31} \[ \int \frac {\left (c x^2\right )^{5/2}}{x^5 (a+b x)} \, dx=\frac {c^2 \sqrt {c x^2} \log (a+b x)}{b x} \]

[In]

Int[(c*x^2)^(5/2)/(x^5*(a + b*x)),x]

[Out]

(c^2*Sqrt[c*x^2]*Log[a + b*x])/(b*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c^2 \sqrt {c x^2}\right ) \int \frac {1}{a+b x} \, dx}{x} \\ & = \frac {c^2 \sqrt {c x^2} \log (a+b x)}{b x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^5 (a+b x)} \, dx=\frac {\left (c x^2\right )^{5/2} \log (a+b x)}{b x^5} \]

[In]

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

[Out]

((c*x^2)^(5/2)*Log[a + b*x])/(b*x^5)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {\left (c \,x^{2}\right )^{\frac {5}{2}} \ln \left (b x +a \right )}{x^{5} b}\)	\(21\)
risch	\(\frac {c^{2} \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b x}\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^5 (a+b x)} \, dx=\frac {\sqrt {c x^{2}} c^{2} \log \left (b x + a\right )}{b x} \]

[In]

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

[Out]

sqrt(c*x^2)*c^2*log(b*x + a)/(b*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.52 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^5 (a+b x)} \, dx=\frac {c^{\frac {5}{2}} \log \left (b x + a\right )}{b} \]

[In]

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

[Out]

c^(5/2)*log(b*x + a)/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {\left (c x^2\right )^{5/2}}{x^5 (a+b x)} \, dx={\left (\frac {c^{2} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b} - \frac {c^{2} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{b}\right )} \sqrt {c} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((c*x^2)^(5/2)/(x^5*(a + b*x)),x)

[Out]

int((c*x^2)^(5/2)/(x^5*(a + b*x)), x)