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

Optimal. Leaf size=106 \[ -\frac{2 (b c-a d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 \sqrt{d}}+\frac{\sqrt{a} (3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{c^2}+\frac{a x \sqrt{a+\frac{b}{x}}}{c} \]

[Out]

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

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Rubi [A]  time = 0.12833, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {375, 98, 156, 63, 208, 205} \[ -\frac{2 (b c-a d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 \sqrt{d}}+\frac{\sqrt{a} (3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{c^2}+\frac{a x \sqrt{a+\frac{b}{x}}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(3/2)/(c + d/x),x]

[Out]

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

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 98

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.19096, size = 102, normalized size = 0.96 \[ \frac{-\frac{2 (b c-a d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{\sqrt{d}}+\sqrt{a} (3 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+a c x \sqrt{a+\frac{b}{x}}}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 528, normalized size = 5. \begin{align*}{\frac{x}{2\,d{c}^{3}}\sqrt{{\frac{ax+b}{x}}} \left ( \ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ) \sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}{b}^{2}{c}^{3}+2\,\sqrt{a}\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}\sqrt{a{x}^{2}+bx}b{c}^{3}+2\,{a}^{3/2}\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}\sqrt{ \left ( ax+b \right ) x}{c}^{2}d-2\,\sqrt{a}\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}\sqrt{ \left ( ax+b \right ) x}b{c}^{3}-2\,{a}^{5/2}\ln \left ({\frac{1}{cx+d} \left ( 2\,\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}\sqrt{ \left ( ax+b \right ) x}c-2\,adx+bcx-bd \right ) } \right ){d}^{3}+4\,{a}^{3/2}\ln \left ({\frac{1}{cx+d} \left ( 2\,\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}\sqrt{ \left ( ax+b \right ) x}c-2\,adx+bcx-bd \right ) } \right ) bc{d}^{2}-2\,\sqrt{a}\ln \left ({\frac{1}{cx+d} \left ( 2\,\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}\sqrt{ \left ( ax+b \right ) x}c-2\,adx+bcx-bd \right ) } \right ){b}^{2}{c}^{2}d-2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}{a}^{2}c{d}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}ab{c}^{2}d-\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ) \sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}{b}^{2}{c}^{3} \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{\frac{ \left ( ad-bc \right ) d}{{c}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x+b)/x)^(1/2)*x*(ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*b^2*c^3
+2*a^(1/2)*((a*d-b*c)*d/c^2)^(1/2)*(a*x^2+b*x)^(1/2)*b*c^3+2*a^(3/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)
*c^2*d-2*a^(1/2)*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*b*c^3-2*a^(5/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x
+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d)/(c*x+d))*d^3+4*a^(3/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a
*d*x+b*c*x-b*d)/(c*x+d))*b*c*d^2-2*a^(1/2)*ln((2*((a*d-b*c)*d/c^2)^(1/2)*((a*x+b)*x)^(1/2)*c-2*a*d*x+b*c*x-b*d
)/(c*x+d))*b^2*c^2*d-2*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*a^2*c*d^2
+3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*a*b*c^2*d-ln(1/2*(2*((a*x+b)*
x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*((a*d-b*c)*d/c^2)^(1/2)*b^2*c^3)/((a*x+b)*x)^(1/2)/d/c^3/a^(1/2)/((a*d-b*c)
*d/c^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x}\right )}^{\frac{3}{2}}}{c + \frac{d}{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76807, size = 1156, normalized size = 10.91 \begin{align*} \left [\frac{2 \, a c x \sqrt{\frac{a x + b}{x}} -{\left (3 \, b c - 2 \, a d\right )} \sqrt{a} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) - 2 \,{\left (b c - a d\right )} \sqrt{-\frac{b c - a d}{d}} \log \left (\frac{2 \, d x \sqrt{-\frac{b c - a d}{d}} \sqrt{\frac{a x + b}{x}} + b d -{\left (b c - 2 \, a d\right )} x}{c x + d}\right )}{2 \, c^{2}}, \frac{a c x \sqrt{\frac{a x + b}{x}} -{\left (3 \, b c - 2 \, a d\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (b c - a d\right )} \sqrt{-\frac{b c - a d}{d}} \log \left (\frac{2 \, d x \sqrt{-\frac{b c - a d}{d}} \sqrt{\frac{a x + b}{x}} + b d -{\left (b c - 2 \, a d\right )} x}{c x + d}\right )}{c^{2}}, \frac{2 \, a c x \sqrt{\frac{a x + b}{x}} + 4 \,{\left (b c - a d\right )} \sqrt{\frac{b c - a d}{d}} \arctan \left (-\frac{d \sqrt{\frac{b c - a d}{d}} \sqrt{\frac{a x + b}{x}}}{b c - a d}\right ) -{\left (3 \, b c - 2 \, a d\right )} \sqrt{a} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right )}{2 \, c^{2}}, \frac{a c x \sqrt{\frac{a x + b}{x}} + 2 \,{\left (b c - a d\right )} \sqrt{\frac{b c - a d}{d}} \arctan \left (-\frac{d \sqrt{\frac{b c - a d}{d}} \sqrt{\frac{a x + b}{x}}}{b c - a d}\right ) -{\left (3 \, b c - 2 \, a d\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right )}{c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError