3.9.91 \(\int \frac {1}{(c x^2)^{3/2} (a+b x)} \, dx\) [891]

Optimal. Leaf size=89 \[ \frac {b}{a^2 c \sqrt {c x^2}}-\frac {1}{2 a c x \sqrt {c x^2}}+\frac {b^2 x \log (x)}{a^3 c \sqrt {c x^2}}-\frac {b^2 x \log (a+b x)}{a^3 c \sqrt {c x^2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {15, 46} \begin {gather*} \frac {b^2 x \log (x)}{a^3 c \sqrt {c x^2}}-\frac {b^2 x \log (a+b x)}{a^3 c \sqrt {c x^2}}+\frac {b}{a^2 c \sqrt {c x^2}}-\frac {1}{2 a c x \sqrt {c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c*x^2)^(3/2)*(a + b*x)),x]

[Out]

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

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

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Mathematica [A]
time = 0.01, size = 51, normalized size = 0.57 \begin {gather*} \frac {x \left (-a (a-2 b x)+2 b^2 x^2 \log (x)-2 b^2 x^2 \log (a+b x)\right )}{2 a^3 \left (c x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x^2)^(3/2)*(a + b*x)),x]

[Out]

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

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Maple [A]
time = 0.11, size = 49, normalized size = 0.55

method	result	size
default	\(\frac {x \left (2 b^{2} \ln \left (x \right ) x^{2}-2 b^{2} \ln \left (b x +a \right ) x^{2}+2 a b x -a^{2}\right )}{2 \left (c \,x^{2}\right )^{\frac {3}{2}} a^{3}}\)	\(49\)
risch	\(\frac {\frac {b x}{a^{2}}-\frac {1}{2 a}}{c x \sqrt {c \,x^{2}}}+\frac {x \,b^{2} \ln \left (-x \right )}{c \sqrt {c \,x^{2}}\, a^{3}}-\frac {b^{2} x \ln \left (b x +a \right )}{a^{3} c \sqrt {c \,x^{2}}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2)^(3/2)/(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 65, normalized size = 0.73 \begin {gather*} -\frac {\left (-1\right )^{\frac {2 \, a c x}{b}} b^{2} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{a^{3} c^{\frac {3}{2}}} + \frac {b}{\sqrt {c x^{2}} a^{2} c} - \frac {1}{2 \, a c^{\frac {3}{2}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2)^(3/2)/(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.45, size = 47, normalized size = 0.53 \begin {gather*} \frac {{\left (2 \, b^{2} x^{2} \log \left (\frac {x}{b x + a}\right ) + 2 \, a b x - a^{2}\right )} \sqrt {c x^{2}}}{2 \, a^{3} c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2)^(3/2)/(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (c x^{2}\right )^{\frac {3}{2}} \left (a + b x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2)**(3/2)/(b*x+a),x)

[Out]

Integral(1/((c*x**2)**(3/2)*(a + b*x)), 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(1/(c*x^2)^(3/2)/(b*x+a),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 {1}{{\left (c\,x^2\right )}^{3/2}\,\left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x^2)^(3/2)*(a + b*x)),x)

[Out]

int(1/((c*x^2)^(3/2)*(a + b*x)), x)

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