\(\int \frac {(c x^2)^{3/2}}{x^7 (a+b x)} \, dx\) [868]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 112 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=-\frac {c \sqrt {c x^2}}{3 a x^4}+\frac {b c \sqrt {c x^2}}{2 a^2 x^3}-\frac {b^2 c \sqrt {c x^2}}{a^3 x^2}-\frac {b^3 c \sqrt {c x^2} \log (x)}{a^4 x}+\frac {b^3 c \sqrt {c x^2} \log (a+b x)}{a^4 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 112, 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, 46} \[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=-\frac {b^3 c \sqrt {c x^2} \log (x)}{a^4 x}+\frac {b^3 c \sqrt {c x^2} \log (a+b x)}{a^4 x}-\frac {b^2 c \sqrt {c x^2}}{a^3 x^2}+\frac {b c \sqrt {c x^2}}{2 a^2 x^3}-\frac {c \sqrt {c x^2}}{3 a x^4} \]

[In]

Int[(c*x^2)^(3/2)/(x^7*(a + b*x)),x]

[Out]

-1/3*(c*Sqrt[c*x^2])/(a*x^4) + (b*c*Sqrt[c*x^2])/(2*a^2*x^3) - (b^2*c*Sqrt[c*x^2])/(a^3*x^2) - (b^3*c*Sqrt[c*x
^2]*Log[x])/(a^4*x) + (b^3*c*Sqrt[c*x^2]*Log[a + b*x])/(a^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 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c \sqrt {c x^2}\right ) \int \frac {1}{x^4 (a+b x)} \, dx}{x} \\ & = \frac {\left (c \sqrt {c x^2}\right ) \int \left (\frac {1}{a x^4}-\frac {b}{a^2 x^3}+\frac {b^2}{a^3 x^2}-\frac {b^3}{a^4 x}+\frac {b^4}{a^4 (a+b x)}\right ) \, dx}{x} \\ & = -\frac {c \sqrt {c x^2}}{3 a x^4}+\frac {b c \sqrt {c x^2}}{2 a^2 x^3}-\frac {b^2 c \sqrt {c x^2}}{a^3 x^2}-\frac {b^3 c \sqrt {c x^2} \log (x)}{a^4 x}+\frac {b^3 c \sqrt {c x^2} \log (a+b x)}{a^4 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.58 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=-\frac {\left (c x^2\right )^{3/2} \left (a \left (2 a^2-3 a b x+6 b^2 x^2\right )+6 b^3 x^3 \log (x)-6 b^3 x^3 \log (a+b x)\right )}{6 a^4 x^6} \]

[In]

Integrate[(c*x^2)^(3/2)/(x^7*(a + b*x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.55

method result size
default \(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a \right ) x^{3}+6 a \,b^{2} x^{2}-3 a^{2} b x +2 a^{3}\right )}{6 x^{6} a^{4}}\) \(62\)
risch \(\frac {c \sqrt {c \,x^{2}}\, \left (-\frac {1}{3 a}+\frac {b x}{2 a^{2}}-\frac {b^{2} x^{2}}{a^{3}}\right )}{x^{4}}-\frac {b^{3} c \ln \left (x \right ) \sqrt {c \,x^{2}}}{a^{4} x}+\frac {c \sqrt {c \,x^{2}}\, b^{3} \ln \left (-b x -a \right )}{x \,a^{4}}\) \(86\)

[In]

int((c*x^2)^(3/2)/x^7/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.53 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=\frac {{\left (6 \, b^{3} c x^{3} \log \left (\frac {b x + a}{x}\right ) - 6 \, a b^{2} c x^{2} + 3 \, a^{2} b c x - 2 \, a^{3} c\right )} \sqrt {c x^{2}}}{6 \, a^{4} x^{4}} \]

[In]

integrate((c*x^2)^(3/2)/x^7/(b*x+a),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=\int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{7} \left (a + b x\right )}\, dx \]

[In]

integrate((c*x**2)**(3/2)/x**7/(b*x+a),x)

[Out]

Integral((c*x**2)**(3/2)/(x**7*(a + b*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=\frac {b^{3} c^{\frac {3}{2}} \log \left (b x + a\right )}{a^{4}} - \frac {b^{3} c^{\frac {3}{2}} \log \left (x\right )}{a^{4}} - \frac {6 \, b^{2} c^{\frac {3}{2}} x^{2} - 3 \, a b c^{\frac {3}{2}} x + 2 \, a^{2} c^{\frac {3}{2}}}{6 \, a^{3} x^{3}} \]

[In]

integrate((c*x^2)^(3/2)/x^7/(b*x+a),x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*x^2)^(3/2)/x^7/(b*x+a),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c x^2\right )^{3/2}}{x^7 (a+b x)} \, dx=\int \frac {{\left (c\,x^2\right )}^{3/2}}{x^7\,\left (a+b\,x\right )} \,d x \]

[In]

int((c*x^2)^(3/2)/(x^7*(a + b*x)),x)

[Out]

int((c*x^2)^(3/2)/(x^7*(a + b*x)), x)