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

Optimal. Leaf size=152 \[ a^{3/2} c (4 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+\frac{c^2 x \left (a+\frac{b}{x}\right )^{7/2}}{a}-\frac{c \left (a+\frac{b}{x}\right )^{5/2} (4 a d+5 b c)}{5 a}-\frac{1}{3} c \left (a+\frac{b}{x}\right )^{3/2} (4 a d+5 b c)-a c \sqrt{a+\frac{b}{x}} (4 a d+5 b c)-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b} \]

[Out]

-(a*c*(5*b*c + 4*a*d)*Sqrt[a + b/x]) - (c*(5*b*c + 4*a*d)*(a + b/x)^(3/2))/3 - (c*(5*b*c + 4*a*d)*(a + b/x)^(5
/2))/(5*a) - (2*d^2*(a + b/x)^(7/2))/(7*b) + (c^2*(a + b/x)^(7/2)*x)/a + a^(3/2)*c*(5*b*c + 4*a*d)*ArcTanh[Sqr
t[a + b/x]/Sqrt[a]]

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Rubi [A]  time = 0.102039, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {375, 89, 80, 50, 63, 208} \[ a^{3/2} c (4 a d+5 b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )+\frac{c^2 x \left (a+\frac{b}{x}\right )^{7/2}}{a}-\frac{c \left (a+\frac{b}{x}\right )^{5/2} (4 a d+5 b c)}{5 a}-\frac{1}{3} c \left (a+\frac{b}{x}\right )^{3/2} (4 a d+5 b c)-a c \sqrt{a+\frac{b}{x}} (4 a d+5 b c)-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^(5/2)*(c + d/x)^2,x]

[Out]

-(a*c*(5*b*c + 4*a*d)*Sqrt[a + b/x]) - (c*(5*b*c + 4*a*d)*(a + b/x)^(3/2))/3 - (c*(5*b*c + 4*a*d)*(a + b/x)^(5
/2))/(5*a) - (2*d^2*(a + b/x)^(7/2))/(7*b) + (c^2*(a + b/x)^(7/2)*x)/a + a^(3/2)*c*(5*b*c + 4*a*d)*ArcTanh[Sqr
t[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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

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

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 )^2 \, dx &=-\operatorname{Subst}\left (\int \frac{(a+b x)^{5/2} (c+d x)^2}{x^2} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c^2 \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{5/2} \left (\frac{1}{2} c (5 b c+4 a d)+a d^2 x\right )}{x} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{(c (5 b c+4 a d)) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=-\frac{c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{1}{2} (c (5 b c+4 a d)) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{1}{3} c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{1}{2} (a c (5 b c+4 a d)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\frac{1}{x}\right )\\ &=-a c (5 b c+4 a d) \sqrt{a+\frac{b}{x}}-\frac{1}{3} c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{1}{2} \left (a^2 c (5 b c+4 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )\\ &=-a c (5 b c+4 a d) \sqrt{a+\frac{b}{x}}-\frac{1}{3} c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{7/2} x}{a}-\frac{\left (a^2 c (5 b c+4 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 c (5 b c+4 a d) \sqrt{a+\frac{b}{x}}-\frac{1}{3} c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{3/2}-\frac{c (5 b c+4 a d) \left (a+\frac{b}{x}\right )^{5/2}}{5 a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b}+\frac{c^2 \left (a+\frac{b}{x}\right )^{7/2} x}{a}+a^{3/2} c (5 b c+4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.135628, size = 121, normalized size = 0.8 \[ -\frac{c (4 a d+5 b c) \left (\sqrt{a+\frac{b}{x}} \left (23 a^2 x^2+11 a b x+3 b^2\right )-15 a^{5/2} x^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )\right )}{15 a x^2}+\frac{c^2 x \left (a+\frac{b}{x}\right )^{7/2}}{a}-\frac{2 d^2 \left (a+\frac{b}{x}\right )^{7/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*(a + b/x)^(7/2))/(7*b) + (c^2*(a + b/x)^(7/2)*x)/a - (c*(5*b*c + 4*a*d)*(Sqrt[a + b/x]*(3*b^2 + 11*a*b
*x + 23*a^2*x^2) - 15*a^(5/2)*x^2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]))/(15*a*x^2)

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Maple [B]  time = 0.013, size = 336, normalized size = 2.2 \begin{align*} -{\frac{1}{210\,b{x}^{4}}\sqrt{{\frac{ax+b}{x}}} \left ( -840\,\sqrt{a{x}^{2}+bx}{a}^{7/2}{x}^{5}cd-1050\,\sqrt{a{x}^{2}+bx}{a}^{5/2}{x}^{5}b{c}^{2}-420\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{5}{a}^{3}bcd-525\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{5}{a}^{2}{b}^{2}{c}^{2}+840\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{5/2}{x}^{3}cd+840\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}{x}^{3}b{c}^{2}+60\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{5/2}{x}^{2}{d}^{2}+448\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}{x}^{2}bcd+140\, \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a}{x}^{2}{b}^{2}{c}^{2}+120\, \left ( a{x}^{2}+bx \right ) ^{3/2}{a}^{3/2}xb{d}^{2}+168\, \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a}x{b}^{2}cd+60\, \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a}{b}^{2}{d}^{2} \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)^2,x)

[Out]

-1/210*((a*x+b)/x)^(1/2)/x^4/b*(-840*(a*x^2+b*x)^(1/2)*a^(7/2)*x^5*c*d-1050*(a*x^2+b*x)^(1/2)*a^(5/2)*x^5*b*c^
2-420*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^5*a^3*b*c*d-525*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1
/2)+2*a*x+b)/a^(1/2))*x^5*a^2*b^2*c^2+840*(a*x^2+b*x)^(3/2)*a^(5/2)*x^3*c*d+840*(a*x^2+b*x)^(3/2)*a^(3/2)*x^3*
b*c^2+60*(a*x^2+b*x)^(3/2)*a^(5/2)*x^2*d^2+448*(a*x^2+b*x)^(3/2)*a^(3/2)*x^2*b*c*d+140*(a*x^2+b*x)^(3/2)*a^(1/
2)*x^2*b^2*c^2+120*(a*x^2+b*x)^(3/2)*a^(3/2)*x*b*d^2+168*(a*x^2+b*x)^(3/2)*a^(1/2)*x*b^2*c*d+60*(a*x^2+b*x)^(3
/2)*a^(1/2)*b^2*d^2)/((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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27295, size = 807, normalized size = 5.31 \begin{align*} \left [\frac{105 \,{\left (5 \, a b^{2} c^{2} + 4 \, a^{2} b c d\right )} \sqrt{a} x^{3} \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (105 \, a^{2} b c^{2} x^{4} - 30 \, b^{3} d^{2} - 2 \,{\left (245 \, a b^{2} c^{2} + 322 \, a^{2} b c d + 15 \, a^{3} d^{2}\right )} x^{3} - 2 \,{\left (35 \, b^{3} c^{2} + 154 \, a b^{2} c d + 45 \, a^{2} b d^{2}\right )} x^{2} - 6 \,{\left (14 \, b^{3} c d + 15 \, a b^{2} d^{2}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{210 \, b x^{3}}, -\frac{105 \,{\left (5 \, a b^{2} c^{2} + 4 \, a^{2} b c d\right )} \sqrt{-a} x^{3} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) -{\left (105 \, a^{2} b c^{2} x^{4} - 30 \, b^{3} d^{2} - 2 \,{\left (245 \, a b^{2} c^{2} + 322 \, a^{2} b c d + 15 \, a^{3} d^{2}\right )} x^{3} - 2 \,{\left (35 \, b^{3} c^{2} + 154 \, a b^{2} c d + 45 \, a^{2} b d^{2}\right )} x^{2} - 6 \,{\left (14 \, b^{3} c d + 15 \, a b^{2} d^{2}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{105 \, b x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/210*(105*(5*a*b^2*c^2 + 4*a^2*b*c*d)*sqrt(a)*x^3*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(105*a^
2*b*c^2*x^4 - 30*b^3*d^2 - 2*(245*a*b^2*c^2 + 322*a^2*b*c*d + 15*a^3*d^2)*x^3 - 2*(35*b^3*c^2 + 154*a*b^2*c*d
+ 45*a^2*b*d^2)*x^2 - 6*(14*b^3*c*d + 15*a*b^2*d^2)*x)*sqrt((a*x + b)/x))/(b*x^3), -1/105*(105*(5*a*b^2*c^2 +
4*a^2*b*c*d)*sqrt(-a)*x^3*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) - (105*a^2*b*c^2*x^4 - 30*b^3*d^2 - 2*(245*a*b^
2*c^2 + 322*a^2*b*c*d + 15*a^3*d^2)*x^3 - 2*(35*b^3*c^2 + 154*a*b^2*c*d + 45*a^2*b*d^2)*x^2 - 6*(14*b^3*c*d +
15*a*b^2*d^2)*x)*sqrt((a*x + b)/x))/(b*x^3)]

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Sympy [A]  time = 57.4778, size = 1841, normalized size = 12.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*a**(19/2)*b**(13/2)*d**2*x**6*sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2)
+ 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) - 40*a**(17/2)*b**(15/2)*d**2*x**5*sqrt(a*x/b + 1
)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**
10*x**(7/2)) - 30*a**(15/2)*b**(17/2)*d**2*x**4*sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*
b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) - 40*a**(13/2)*b**(19/2)*d**2*x**3*
sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 10
5*a**(7/2)*b**10*x**(7/2)) + 8*a**(13/2)*b**(5/2)*d**2*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)) - 100*a**(11/2)*b**(21/2)*d**2*x**2*sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*
a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) + 8*a**(11/2)*b**(7/2)*c*
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)) + 4*a**(11/2)*b**(7/2)*d**2*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)) - 96*a**(9/2)*b**(23/2)*d**2*x*sqrt(
a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**
(7/2)*b**10*x**(7/2)) + 4*a**(9/2)*b**(9/2)*c*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)) - 16*a**(9/2)*b**(9/2)*d**2*x*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**
(5/2)) - 30*a**(7/2)*b**(25/2)*d**2*sqrt(a*x/b + 1)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/
2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) - 16*a**(7/2)*b**(11/2)*c*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)) - 12*a**(7/2)*b**(11/2)*d**2*sqrt(a*x/b + 1)/(15*a**(7/
2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 12*a**(5/2)*b**(13/2)*c*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**2*asinh(sqrt(a)*sqrt(x)/sqrt(b)) + 16*a**10*b**6*d**2*x**(
13/2)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)
*b**10*x**(7/2)) + 48*a**9*b**7*d**2*x**(11/2)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) +
315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) + 48*a**8*b**8*d**2*x**(9/2)/(105*a**(13/2)*b**7*x**
(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 105*a**(7/2)*b**10*x**(7/2)) + 16*a**7*b*
*9*d**2*x**(7/2)/(105*a**(13/2)*b**7*x**(13/2) + 315*a**(11/2)*b**8*x**(11/2) + 315*a**(9/2)*b**9*x**(9/2) + 1
05*a**(7/2)*b**10*x**(7/2)) - 8*a**7*b**2*d**2*x**(7/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)
) - 8*a**6*b**3*c*d*x**(7/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 8*a**6*b**3*d**2*x**(5/
2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 8*a**5*b**4*c*d*x**(5/2)/(15*a**(7/2)*b**3*x**(7/
2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**3*c*d*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) + a**2*sqrt(b)*c**2*sqrt(x)
*sqrt(a*x/b + 1) - 4*a**2*b*c**2*atan(sqrt(a + b/x)/sqrt(-a))/sqrt(-a) - 4*a**2*c*d*sqrt(a + b/x) + a**2*d**2*
Piecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True)) - 4*a*b*c**2*sqrt(a + b/x) + 4*a*b*c*d*Pi
ecewise((-sqrt(a)/x, Eq(b, 0)), (-2*(a + b/x)**(3/2)/(3*b), True)) + b**2*c**2*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)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError