3.10.3 \(\int \frac {(c x^2)^{3/2}}{x (a+b x)^2} \, dx\) [903]

Optimal. Leaf size=68 \[ \frac {c \sqrt {c x^2}}{b^2}-\frac {a^2 c \sqrt {c x^2}}{b^3 x (a+b x)}-\frac {2 a c \sqrt {c x^2} \log (a+b x)}{b^3 x} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 68, 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, 45} \begin {gather*} -\frac {a^2 c \sqrt {c x^2}}{b^3 x (a+b x)}-\frac {2 a c \sqrt {c x^2} \log (a+b x)}{b^3 x}+\frac {c \sqrt {c x^2}}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x^2)^(3/2)/(x*(a + b*x)^2),x]

[Out]

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

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Mathematica [A]
time = 0.01, size = 55, normalized size = 0.81 \begin {gather*} \frac {c^2 x \left (-a^2+a b x+b^2 x^2-2 a (a+b x) \log (a+b x)\right )}{b^3 \sqrt {c x^2} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 62, normalized size = 0.91

method	result	size
default	\(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}} \left (2 \ln \left (b x +a \right ) a b x -x^{2} b^{2}+2 a^{2} \ln \left (b x +a \right )-a b x +a^{2}\right )}{x^{3} b^{3} \left (b x +a \right )}\)	\(62\)
risch	\(\frac {c \sqrt {c \,x^{2}}}{b^{2}}-\frac {a^{2} c \sqrt {c \,x^{2}}}{b^{3} x \left (b x +a \right )}-\frac {2 a c \ln \left (b x +a \right ) \sqrt {c \,x^{2}}}{b^{3} x}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 98, normalized size = 1.44 \begin {gather*} \frac {\sqrt {c x^{2}} a c}{b^{3} x + a b^{2}} - \frac {2 \, \left (-1\right )^{\frac {2 \, c x}{b}} a c^{\frac {3}{2}} \log \left (\frac {2 \, c x}{b}\right )}{b^{3}} - \frac {2 \, \left (-1\right )^{\frac {2 \, a c x}{b}} a c^{\frac {3}{2}} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b^{3}} + \frac {\sqrt {c x^{2}} c}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.46, size = 63, normalized size = 0.93 \begin {gather*} \frac {{\left (b^{2} c x^{2} + a b c x - a^{2} c - 2 \, {\left (a b c x + a^{2} c\right )} \log \left (b x + a\right )\right )} \sqrt {c x^{2}}}{b^{4} x^{2} + a b^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 \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/(b*x+a)**2,x)

[Out]

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

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Giac [A]
time = 0.54, size = 58, normalized size = 0.85 \begin {gather*} c^{\frac {3}{2}} {\left (\frac {x \mathrm {sgn}\left (x\right )}{b^{2}} - \frac {2 \, a \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {{\left (2 \, a \log \left ({\left | a \right |}\right ) + a\right )} \mathrm {sgn}\left (x\right )}{b^{3}} - \frac {a^{2} \mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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\,{\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*(a + b*x)^2),x)

[Out]

int((c*x^2)^(3/2)/(x*(a + b*x)^2), x)

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