3.10.7 \(\int \frac {(c x^2)^{3/2}}{x^5 (a+b x)^2} \, dx\) [907]

Optimal. Leaf size=91 \[ -\frac {c \sqrt {c x^2}}{a^2 x^2}-\frac {b c \sqrt {c x^2}}{a^2 x (a+b x)}-\frac {2 b c \sqrt {c x^2} \log (x)}{a^3 x}+\frac {2 b c \sqrt {c x^2} \log (a+b x)}{a^3 x} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 46} \begin {gather*} -\frac {2 b c \sqrt {c x^2} \log (x)}{a^3 x}+\frac {2 b c \sqrt {c x^2} \log (a+b x)}{a^3 x}-\frac {b c \sqrt {c x^2}}{a^2 x (a+b x)}-\frac {c \sqrt {c x^2}}{a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(3/2)/(x^5*(a + b*x)^2),x]

[Out]

-((c*Sqrt[c*x^2])/(a^2*x^2)) - (b*c*Sqrt[c*x^2])/(a^2*x*(a + b*x)) - (2*b*c*Sqrt[c*x^2]*Log[x])/(a^3*x) + (2*b
*c*Sqrt[c*x^2]*Log[a + b*x])/(a^3*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*} \int \frac {\left (c x^2\right )^{3/2}}{x^5 (a+b x)^2} \, dx &=\frac {\left (c \sqrt {c x^2}\right ) \int \frac {1}{x^2 (a+b x)^2} \, dx}{x}\\ &=\frac {\left (c \sqrt {c x^2}\right ) \int \left (\frac {1}{a^2 x^2}-\frac {2 b}{a^3 x}+\frac {b^2}{a^2 (a+b x)^2}+\frac {2 b^2}{a^3 (a+b x)}\right ) \, dx}{x}\\ &=-\frac {c \sqrt {c x^2}}{a^2 x^2}-\frac {b c \sqrt {c x^2}}{a^2 x (a+b x)}-\frac {2 b c \sqrt {c x^2} \log (x)}{a^3 x}+\frac {2 b c \sqrt {c x^2} \log (a+b x)}{a^3 x}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 59, normalized size = 0.65 \begin {gather*} -\frac {c^2 (a (a+2 b x)+2 b x (a+b x) \log (x)-2 b x (a+b x) \log (a+b x))}{a^3 \sqrt {c x^2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x^2)^(3/2)/(x^5*(a + b*x)^2),x]

[Out]

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

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Maple [A]
time = 0.13, size = 74, normalized size = 0.81

method	result	size
default	\(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (2 b^{2} \ln \left (x \right ) x^{2}-2 b^{2} \ln \left (b x +a \right ) x^{2}+2 a b \ln \left (x \right ) x -2 \ln \left (b x +a \right ) a b x +2 a b x +a^{2}\right )}{x^{4} a^{3} \left (b x +a \right )}\)	\(74\)
risch	\(\frac {c \sqrt {c \,x^{2}}\, \left (-\frac {2 b x}{a^{2}}-\frac {1}{a}\right )}{x^{2} \left (b x +a \right )}-\frac {2 b c \ln \left (x \right ) \sqrt {c \,x^{2}}}{a^{3} x}+\frac {2 c \sqrt {c \,x^{2}}\, b \ln \left (-b x -a \right )}{x \,a^{3}}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(3/2)/x^5/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 58, normalized size = 0.64 \begin {gather*} \frac {2 \, b c^{\frac {3}{2}} \log \left (b x + a\right )}{a^{3}} - \frac {2 \, b c^{\frac {3}{2}} \log \left (x\right )}{a^{3}} - \frac {2 \, b c^{\frac {3}{2}} x + a c^{\frac {3}{2}}}{a^{2} b x^{2} + a^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)/x^5/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.43, size = 65, normalized size = 0.71 \begin {gather*} -\frac {{\left (2 \, a b c x + a^{2} c - 2 \, {\left (b^{2} c x^{2} + a b c x\right )} \log \left (\frac {b x + a}{x}\right )\right )} \sqrt {c x^{2}}}{a^{3} b x^{3} + a^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)/x^5/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{x^{5} \left (a + b x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2)**(3/2)/x**5/(b*x+a)**2,x)

[Out]

Integral((c*x**2)**(3/2)/(x**5*(a + b*x)**2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2)^(3/2)/x^5/(b*x+a)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2\right )}^{3/2}}{x^5\,{\left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2)^(3/2)/(x^5*(a + b*x)^2),x)

[Out]

int((c*x^2)^(3/2)/(x^5*(a + b*x)^2), x)

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