3.242 \(\int \frac{(a+\frac{b}{x})^{5/2}}{c+\frac{d}{x}} \, dx\)

Optimal. Leaf size=134 \[ \frac{a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{c^2}+\frac{2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 d^{3/2}}-\frac{b \sqrt{a+\frac{b}{x}} (a d+2 b c)}{c d}+\frac{a x \left (a+\frac{b}{x}\right )^{3/2}}{c} \]

[Out]

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

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Rubi [A]  time = 0.221343, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {375, 98, 154, 156, 63, 208, 205} \[ \frac{a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{c^2}+\frac{2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 d^{3/2}}-\frac{b \sqrt{a+\frac{b}{x}} (a d+2 b c)}{c d}+\frac{a x \left (a+\frac{b}{x}\right )^{3/2}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)/(c + d/x),x]

[Out]

-((b*(2*b*c + a*d)*Sqrt[a + b/x])/(c*d)) + (a*(a + b/x)^(3/2)*x)/c + (2*(b*c - a*d)^(5/2)*ArcTan[(Sqrt[d]*Sqrt
[a + b/x])/Sqrt[b*c - a*d]])/(c^2*d^(3/2)) + (a^(3/2)*(5*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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

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 )^{5/2}}{c+\frac{d}{x}} \, dx &=-\operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^2 (c+d x)} \, dx,x,\frac{1}{x}\right )\\ &=\frac{a \left (a+\frac{b}{x}\right )^{3/2} x}{c}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x} \left (-\frac{1}{2} a (5 b c-2 a d)-\frac{1}{2} b (2 b c+a d) x\right )}{x (c+d x)} \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{b (2 b c+a d) \sqrt{a+\frac{b}{x}}}{c d}+\frac{a \left (a+\frac{b}{x}\right )^{3/2} x}{c}+\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{1}{4} a^2 d (5 b c-2 a d)+\frac{1}{4} b \left (2 b^2 c^2-6 a b c d+a^2 d^2\right ) x}{x \sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{c d}\\ &=-\frac{b (2 b c+a d) \sqrt{a+\frac{b}{x}}}{c d}+\frac{a \left (a+\frac{b}{x}\right )^{3/2} x}{c}-\frac{\left (a^2 (5 b c-2 a d)\right ) \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)^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x} (c+d x)} \, dx,x,\frac{1}{x}\right )}{c^2 d}\\ &=-\frac{b (2 b c+a d) \sqrt{a+\frac{b}{x}}}{c d}+\frac{a \left (a+\frac{b}{x}\right )^{3/2} x}{c}-\frac{\left (a^2 (5 b c-2 a d)\right ) \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)^3\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 d}\\ &=-\frac{b (2 b c+a d) \sqrt{a+\frac{b}{x}}}{c d}+\frac{a \left (a+\frac{b}{x}\right )^{3/2} x}{c}+\frac{2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{c^2 d^{3/2}}+\frac{a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{c^2}\\ \end{align*}

Mathematica [A]  time = 0.183027, size = 116, normalized size = 0.87 \[ \frac{\frac{c \sqrt{a+\frac{b}{x}} \left (a^2 d x-2 b^2 c\right )}{d}+a^{3/2} (5 b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+\frac{2 (b c-a d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{a+\frac{b}{x}}}{\sqrt{b c-a d}}\right )}{d^{3/2}}}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^(5/2)/(c + d/x),x]

[Out]

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

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Maple [B]  time = 0.013, size = 859, normalized size = 6.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^(5/2)/(c+d/x),x)

[Out]

-1/2*((a*x+b)/x)^(1/2)/x*(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)*x^2*
a^3*c*d^3-5*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)*x^2*a^2*b*c^2*d^2+4*
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)*x^2*a*b^2*c^3*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)*x^2*b^3*c^4-8*((a*d-b*c)*d/c^2)^(1/2)*(a*x^2+b*x
)^(1/2)*a^(3/2)*x^2*b*c^3*d+2*((a*d-b*c)*d/c^2)^(1/2)*(a*x^2+b*x)^(1/2)*a^(1/2)*x^2*b^2*c^4-4*((a*d-b*c)*d/c^2
)^(1/2)*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a*b^2*c^3*d+((a*d-b*c)*d/c^2)^(1/2)*ln(1/2*(
2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*b^3*c^4-2*((a*d-b*c)*d/c^2)^(1/2)*a^(5/2)*((a*x+b)*x)^(1/2)*
x^2*c^2*d^2+4*((a*d-b*c)*d/c^2)^(1/2)*a^(3/2)*((a*x+b)*x)^(1/2)*x^2*b*c^3*d-2*((a*d-b*c)*d/c^2)^(1/2)*a^(1/2)*
((a*x+b)*x)^(1/2)*x^2*b^2*c^4+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))*
a^(7/2)*x^2*d^4-6*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))*a^(5/2)*x^2*b*
c*d^3+6*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))*a^(3/2)*x^2*b^2*c^2*d^2-
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))*a^(1/2)*x^2*b^3*c^3*d+4*((a*d-
b*c)*d/c^2)^(1/2)*(a*x^2+b*x)^(3/2)*a^(1/2)*b*c^3*d)/((a*x+b)*x)^(1/2)/d^2/a^(1/2)/c^3/((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{5}{2}}}{c + \frac{d}{x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/(c+d/x),x, algorithm="maxima")

[Out]

integrate((a + b/x)^(5/2)/(c + d/x), x)

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Fricas [A]  time = 2.26667, size = 1431, normalized size = 10.68 \begin{align*} \left [-\frac{{\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt{a} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) - 2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 \,{\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt{\frac{a x + b}{x}}}{2 \, c^{2} d}, -\frac{{\left (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 ) -{\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt{\frac{a x + b}{x}}}{c^{2} d}, -\frac{4 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt{a} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) - 2 \,{\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt{\frac{a x + b}{x}}}{2 \, c^{2} d}, -\frac{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\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 (5 \, a b c d - 2 \, a^{2} d^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (a^{2} c d x - 2 \, b^{2} c^{2}\right )} \sqrt{\frac{a x + b}{x}}}{c^{2} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^(5/2)/(c+d/x),x, algorithm="fricas")

[Out]

[-1/2*((5*a*b*c*d - 2*a^2*d^2)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(b^2*c^2 - 2*a*b*c*d
+ a^2*d^2)*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)) - 2*(a^2*c*d*x - 2*b^2*c^2)*sqrt((a*x + b)/x))/(c^2*d), -((5*a*b*c*d - 2*a^2*d^2)*sqrt(-a)*arctan(sqr
t(-a)*sqrt((a*x + b)/x)/a) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*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)) - (a^2*c*d*x - 2*b^2*c^2)*sqrt((a*x + b)/x))/(c^2*d)
, -1/2*(4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt((b*c - a*d)/d)*arctan(-d*sqrt((b*c - a*d)/d)*sqrt((a*x + b)/x)/
(b*c - a*d)) + (5*a*b*c*d - 2*a^2*d^2)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(a^2*c*d*x -
2*b^2*c^2)*sqrt((a*x + b)/x))/(c^2*d), -(2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt((b*c - a*d)/d)*arctan(-d*sqrt
((b*c - a*d)/d)*sqrt((a*x + b)/x)/(b*c - a*d)) + (5*a*b*c*d - 2*a^2*d^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x +
b)/x)/a) - (a^2*c*d*x - 2*b^2*c^2)*sqrt((a*x + b)/x))/(c^2*d)]

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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{5}{2}}}{c x + d}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**(5/2)/(c+d/x),x)

[Out]

Integral(x*(a + b/x)**(5/2)/(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)^(5/2)/(c+d/x),x, algorithm="giac")

[Out]

Exception raised: TypeError