3.764 \(\int \frac{(c+d x)^{3/2}}{x^2 (a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=108 \[ -\frac{3 \sqrt{c+d x} (b c-a d)}{a^2 \sqrt{a+b x}}+\frac{3 \sqrt{c} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}}-\frac{(c+d x)^{3/2}}{a x \sqrt{a+b x}} \]

[Out]

(-3*(b*c - a*d)*Sqrt[c + d*x])/(a^2*Sqrt[a + b*x]) - (c + d*x)^(3/2)/(a*x*Sqrt[a + b*x]) + (3*Sqrt[c]*(b*c - a
*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.0402751, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {94, 93, 208} \[ -\frac{3 \sqrt{c+d x} (b c-a d)}{a^2 \sqrt{a+b x}}+\frac{3 \sqrt{c} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}}-\frac{(c+d x)^{3/2}}{a x \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*c - a*d)*Sqrt[c + d*x])/(a^2*Sqrt[a + b*x]) - (c + d*x)^(3/2)/(a*x*Sqrt[a + b*x]) + (3*Sqrt[c]*(b*c - a
*d)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2)

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(c+d x)^{3/2}}{x^2 (a+b x)^{3/2}} \, dx &=-\frac{(c+d x)^{3/2}}{a x \sqrt{a+b x}}-\frac{(3 (b c-a d)) \int \frac{\sqrt{c+d x}}{x (a+b x)^{3/2}} \, dx}{2 a}\\ &=-\frac{3 (b c-a d) \sqrt{c+d x}}{a^2 \sqrt{a+b x}}-\frac{(c+d x)^{3/2}}{a x \sqrt{a+b x}}-\frac{(3 c (b c-a d)) \int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx}{2 a^2}\\ &=-\frac{3 (b c-a d) \sqrt{c+d x}}{a^2 \sqrt{a+b x}}-\frac{(c+d x)^{3/2}}{a x \sqrt{a+b x}}-\frac{(3 c (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{c+d x}}\right )}{a^2}\\ &=-\frac{3 (b c-a d) \sqrt{c+d x}}{a^2 \sqrt{a+b x}}-\frac{(c+d x)^{3/2}}{a x \sqrt{a+b x}}+\frac{3 \sqrt{c} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0824342, size = 91, normalized size = 0.84 \[ \frac{\sqrt{c+d x} (-a c+2 a d x-3 b c x)}{a^2 x \sqrt{a+b x}}+\frac{3 \sqrt{c} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{a^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x]*(-(a*c) - 3*b*c*x + 2*a*d*x))/(a^2*x*Sqrt[a + b*x]) + (3*Sqrt[c]*(b*c - a*d)*ArcTanh[(Sqrt[c]*S
qrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/a^(5/2)

________________________________________________________________________________________

Maple [B]  time = 0.021, size = 298, normalized size = 2.8 \begin{align*} -{\frac{1}{2\,{a}^{2}x}\sqrt{dx+c} \left ( 3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}abcd-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ){x}^{2}{b}^{2}{c}^{2}+3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) x{a}^{2}cd-3\,\ln \left ({\frac{adx+bcx+2\,\sqrt{ac}\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }+2\,ac}{x}} \right ) xab{c}^{2}-4\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xad+6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xbc+2\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }ac\sqrt{ac} \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 3.61694, size = 752, normalized size = 6.96 \begin{align*} \left [-\frac{3 \,{\left ({\left (b^{2} c - a b d\right )} x^{2} +{\left (a b c - a^{2} d\right )} x\right )} \sqrt{\frac{c}{a}} \log \left (\frac{8 \, a^{2} c^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \,{\left (2 \, a^{2} c +{\left (a b c + a^{2} d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{\frac{c}{a}} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \,{\left (a c +{\left (3 \, b c - 2 \, a d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{4 \,{\left (a^{2} b x^{2} + a^{3} x\right )}}, -\frac{3 \,{\left ({\left (b^{2} c - a b d\right )} x^{2} +{\left (a b c - a^{2} d\right )} x\right )} \sqrt{-\frac{c}{a}} \arctan \left (\frac{{\left (2 \, a c +{\left (b c + a d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{-\frac{c}{a}}}{2 \,{\left (b c d x^{2} + a c^{2} +{\left (b c^{2} + a c d\right )} x\right )}}\right ) + 2 \,{\left (a c +{\left (3 \, b c - 2 \, a d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (a^{2} b x^{2} + a^{3} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(3*((b^2*c - a*b*d)*x^2 + (a*b*c - a^2*d)*x)*sqrt(c/a)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*
x^2 - 4*(2*a^2*c + (a*b*c + a^2*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(c/a) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) +
4*(a*c + (3*b*c - 2*a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b*x^2 + a^3*x), -1/2*(3*((b^2*c - a*b*d)*x^2 + (
a*b*c - a^2*d)*x)*sqrt(-c/a)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-c/a)/(b*c*d*
x^2 + a*c^2 + (b*c^2 + a*c*d)*x)) + 2*(a*c + (3*b*c - 2*a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b*x^2 + a^3*
x)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError