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

Optimal. Leaf size=125 \[ a^{3/2} (2 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )-\frac{\left (a+\frac{b}{x}\right )^{5/2} (2 a d+5 b c)}{5 a}-\frac{1}{3} \left (a+\frac{b}{x}\right )^{3/2} (2 a d+5 b c)-a \sqrt{a+\frac{b}{x}} (2 a d+5 b c)+\frac{c x \left (a+\frac{b}{x}\right )^{7/2}}{a} \]

[Out]

-(a*(5*b*c + 2*a*d)*Sqrt[a + b/x]) - ((5*b*c + 2*a*d)*(a + b/x)^(3/2))/3 - ((5*b*c + 2*a*d)*(a + b/x)^(5/2))/(
5*a) + (c*(a + b/x)^(7/2)*x)/a + a^(3/2)*(5*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

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Rubi [A]  time = 0.0766491, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {375, 78, 50, 63, 208} \[ a^{3/2} (2 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )-\frac{\left (a+\frac{b}{x}\right )^{5/2} (2 a d+5 b c)}{5 a}-\frac{1}{3} \left (a+\frac{b}{x}\right )^{3/2} (2 a d+5 b c)-a \sqrt{a+\frac{b}{x}} (2 a d+5 b c)+\frac{c x \left (a+\frac{b}{x}\right )^{7/2}}{a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*(c + d/x),x]

[Out]

-(a*(5*b*c + 2*a*d)*Sqrt[a + b/x]) - ((5*b*c + 2*a*d)*(a + b/x)^(3/2))/3 - ((5*b*c + 2*a*d)*(a + b/x)^(5/2))/(
5*a) + (c*(a + b/x)^(7/2)*x)/a + a^(3/2)*(5*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

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 78

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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]

Rubi steps

\begin{align*} \int \left (a+\frac{b}{x}\right )^{5/2} \left (c+\frac{d}{x}\right ) \, dx &=-\operatorname{Subst}\left (\int \frac{(a+b x)^{5/2} (c+d x)}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{\left (\frac{5 b c}{2}+a d\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{(5 b c+2 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}+\frac{c \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{1}{2} (5 b c+2 a d) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{1}{3} (5 b c+2 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{(5 b c+2 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}+\frac{c \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{1}{2} (a (5 b c+2 a d)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\frac{1}{x}\right )\\ &=-a (5 b c+2 a d) \sqrt{a+\frac{b}{x}}-\frac{1}{3} (5 b c+2 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{(5 b c+2 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}+\frac{c \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{1}{2} \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 )\\ &=-a (5 b c+2 a d) \sqrt{a+\frac{b}{x}}-\frac{1}{3} (5 b c+2 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{(5 b c+2 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}+\frac{c \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\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}\\ &=-a (5 b c+2 a d) \sqrt{a+\frac{b}{x}}-\frac{1}{3} (5 b c+2 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{(5 b c+2 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}+\frac{c \left (a+\frac{b}{x}\right )^{7/2} x}{a}+a^{3/2} (5 b c+2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0938498, size = 94, normalized size = 0.75 \[ \frac{\sqrt{a+\frac{b}{x}} \left (a^2 x^2 (15 c x-46 d)-2 a b x (35 c x+11 d)-2 b^2 (5 c x+3 d)\right )}{15 x^2}+a^{3/2} (2 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b/x]*(-2*b^2*(3*d + 5*c*x) + a^2*x^2*(-46*d + 15*c*x) - 2*a*b*x*(11*d + 35*c*x)))/(15*x^2) + a^(3/2)
*(5*b*c + 2*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

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Maple [B]  time = 0.01, size = 253, normalized size = 2. \begin{align*} -{\frac{1}{30\,b{x}^{3}}\sqrt{{\frac{ax+b}{x}}} \left ( -60\,\sqrt{a{x}^{2}+bx}{a}^{7/2}{x}^{4}d-150\,\sqrt{a{x}^{2}+bx}{a}^{5/2}{x}^{4}bc-30\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{4}{a}^{3}bd-75\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{4}{a}^{2}{b}^{2}c+60\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{5/2}{x}^{2}d+120\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}{x}^{2}bc+32\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}xbd+20\, \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a}x{b}^{2}c+12\, \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a}{b}^{2}d \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*((a*x+b)/x)^(1/2)/x^3/b*(-60*(a*x^2+b*x)^(1/2)*a^(7/2)*x^4*d-150*(a*x^2+b*x)^(1/2)*a^(5/2)*x^4*b*c-30*ln
(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^4*a^3*b*d-75*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b
)/a^(1/2))*x^4*a^2*b^2*c+60*(a*x^2+b*x)^(3/2)*a^(5/2)*x^2*d+120*(a*x^2+b*x)^(3/2)*a^(3/2)*x^2*b*c+32*(a*x^2+b*
x)^(3/2)*a^(3/2)*x*b*d+20*(a*x^2+b*x)^(3/2)*a^(1/2)*x*b^2*c+12*(a*x^2+b*x)^(3/2)*a^(1/2)*b^2*d)/((a*x+b)*x)^(1
/2)/a^(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((a+b/x)^(5/2)*(c+d/x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.23765, size = 527, normalized size = 4.22 \begin{align*} \left [\frac{15 \,{\left (5 \, a b c + 2 \, a^{2} d\right )} \sqrt{a} x^{2} \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (15 \, a^{2} c x^{3} - 6 \, b^{2} d - 2 \,{\left (35 \, a b c + 23 \, a^{2} d\right )} x^{2} - 2 \,{\left (5 \, b^{2} c + 11 \, a b d\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{30 \, x^{2}}, -\frac{15 \,{\left (5 \, a b c + 2 \, a^{2} d\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (15 \, a^{2} c x^{3} - 6 \, b^{2} d - 2 \,{\left (35 \, a b c + 23 \, a^{2} d\right )} x^{2} - 2 \,{\left (5 \, b^{2} c + 11 \, a b d\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{15 \, x^{2}}\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]

[1/30*(15*(5*a*b*c + 2*a^2*d)*sqrt(a)*x^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(15*a^2*c*x^3 - 6
*b^2*d - 2*(35*a*b*c + 23*a^2*d)*x^2 - 2*(5*b^2*c + 11*a*b*d)*x)*sqrt((a*x + b)/x))/x^2, -1/15*(15*(5*a*b*c +
2*a^2*d)*sqrt(-a)*x^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (15*a^2*c*x^3 - 6*b^2*d - 2*(35*a*b*c + 23*a^2*d)
*x^2 - 2*(5*b^2*c + 11*a*b*d)*x)*sqrt((a*x + b)/x))/x^2]

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Sympy [A]  time = 34.9096, size = 520, normalized size = 4.16 \begin{align*} \frac{4 a^{\frac{11}{2}} b^{\frac{7}{2}} d x^{3} \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} + \frac{2 a^{\frac{9}{2}} b^{\frac{9}{2}} d x^{2} \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{8 a^{\frac{7}{2}} b^{\frac{11}{2}} d x \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{6 a^{\frac{5}{2}} b^{\frac{13}{2}} d \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} + a^{\frac{3}{2}} b c \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )} - \frac{4 a^{6} b^{3} d x^{\frac{7}{2}}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{4 a^{5} b^{4} d x^{\frac{5}{2}}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{2 a^{3} d \operatorname{atan}{\left (\frac{\sqrt{a + \frac{b}{x}}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + a^{2} \sqrt{b} c \sqrt{x} \sqrt{\frac{a x}{b} + 1} - \frac{4 a^{2} b c \operatorname{atan}{\left (\frac{\sqrt{a + \frac{b}{x}}}{\sqrt{- a}} \right )}}{\sqrt{- a}} - 2 a^{2} d \sqrt{a + \frac{b}{x}} - 4 a b c \sqrt{a + \frac{b}{x}} + 2 a b d \left (\begin{cases} - \frac{\sqrt{a}}{x} & \text{for}\: b = 0 \\- \frac{2 \left (a + \frac{b}{x}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) + b^{2} c \left (\begin{cases} - \frac{\sqrt{a}}{x} & \text{for}\: b = 0 \\- \frac{2 \left (a + \frac{b}{x}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a**(11/2)*b**(7/2)*d*x**3*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + 2*a**(9/
2)*b**(9/2)*d*x**2*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 8*a**(7/2)*b**(11
/2)*d*x*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 6*a**(5/2)*b**(13/2)*d*sqrt(
a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + a**(3/2)*b*c*asinh(sqrt(a)*sqrt(x)/sqrt(b
)) - 4*a**6*b**3*d*x**(7/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**5*b**4*d*x**(5/2)/(
15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 2*a**3*d*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) + a**2
*sqrt(b)*c*sqrt(x)*sqrt(a*x/b + 1) - 4*a**2*b*c*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) - 2*a**2*d*sqrt(a + b/x)
- 4*a*b*c*sqrt(a + b/x) + 2*a*b*d*Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True)) + b**2
*c*Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True))

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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