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

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

[Out]

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

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Rubi [A]  time = 0.0400652, 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} \[ \frac{3 \sqrt{a+b x} (b c-a d)}{c^2 \sqrt{c+d x}}-\frac{3 \sqrt{a} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}}-\frac{(a+b x)^{3/2}}{c x \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.07891, size = 91, normalized size = 0.84 \[ \frac{\sqrt{a+b x} (2 b c x-a (c+3 d x))}{c^2 x \sqrt{c+d x}}+\frac{3 \sqrt{a} (a d-b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{c+d x}}\right )}{c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.021, size = 298, normalized size = 2.8 \begin{align*}{\frac{1}{2\,{c}^{2}x}\sqrt{bx+a} \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}{a}^{2}{d}^{2}-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{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}-6\,\sqrt{ \left ( bx+a \right ) \left ( dx+c \right ) }\sqrt{ac}xad+4\,\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{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.57048, size = 751, normalized size = 6.95 \begin{align*} \left [-\frac{3 \,{\left ({\left (b c d - a d^{2}\right )} x^{2} +{\left (b c^{2} - a c d\right )} x\right )} \sqrt{\frac{a}{c}} \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 c^{2} +{\left (b c^{2} + a c d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c} \sqrt{\frac{a}{c}} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) + 4 \,{\left (a c -{\left (2 \, b c - 3 \, a d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{4 \,{\left (c^{2} d x^{2} + c^{3} x\right )}}, \frac{3 \,{\left ({\left (b c d - a d^{2}\right )} x^{2} +{\left (b c^{2} - a c d\right )} x\right )} \sqrt{-\frac{a}{c}} \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{a}{c}}}{2 \,{\left (a b d x^{2} + a^{2} c +{\left (a b c + a^{2} d\right )} x\right )}}\right ) - 2 \,{\left (a c -{\left (2 \, b c - 3 \, a d\right )} x\right )} \sqrt{b x + a} \sqrt{d x + c}}{2 \,{\left (c^{2} d x^{2} + c^{3} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Exception raised: TypeError