3.3.24 \(\int \sqrt {a+\frac {b}{x}} (c+\frac {d}{x})^3 \, dx\) [224]

Optimal. Leaf size=143 \[ -\frac {7}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (57 b^2 c^2+15 a b c d-2 a^2 d^2\right )+\frac {b d (33 b c+2 a d)}{x}\right )}{15 b^2}+\sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x+\frac {c^2 (b c+6 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

c^2*(6*a*d+b*c)*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(1/2)-7/5*d*(c+d/x)^2*(a+b/x)^(1/2)-1/15*d*(-4*a^2*d^2+30*a*b
*c*d+114*b^2*c^2+b*d*(2*a*d+33*b*c)/x)*(a+b/x)^(1/2)/b^2+(c+d/x)^3*x*(a+b/x)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {382, 99, 158, 152, 65, 214} \begin {gather*} -\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (-2 a^2 d^2+15 a b c d+57 b^2 c^2\right )+\frac {b d (2 a d+33 b c)}{x}\right )}{15 b^2}+\frac {c^2 (6 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}}+x \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3-\frac {7}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]*(c + d/x)^3,x]

[Out]

(-7*d*Sqrt[a + b/x]*(c + d/x)^2)/5 - (d*Sqrt[a + b/x]*(2*(57*b^2*c^2 + 15*a*b*c*d - 2*a^2*d^2) + (b*d*(33*b*c
+ 2*a*d))/x))/(15*b^2) + Sqrt[a + b/x]*(c + d/x)^3*x + (c^2*(b*c + 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/Sqrt
[a]

Rule 65

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 99

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/(b*(m + 1))), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)
^(m + 1)*((c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d
*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1
)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 158

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] && IntegerQ[m]

Rule 214

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

Rule 382

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]

Rubi steps

\begin {align*} \int \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 \, dx &=-\text {Subst}\left (\int \frac {\sqrt {a+b x} (c+d x)^3}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x-\text {Subst}\left (\int \frac {(c+d x)^2 \left (\frac {1}{2} (b c+6 a d)+\frac {7 b d x}{2}\right )}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {7}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2+\sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x-\frac {2 \text {Subst}\left (\int \frac {(c+d x) \left (\frac {5}{4} b c (b c+6 a d)+\frac {1}{4} b d (33 b c+2 a d) x\right )}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{5 b}\\ &=-\frac {7}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (57 b^2 c^2+15 a b c d-2 a^2 d^2\right )+\frac {b d (33 b c+2 a d)}{x}\right )}{15 b^2}+\sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x-\frac {1}{2} \left (c^2 (b c+6 a d)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {7}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (57 b^2 c^2+15 a b c d-2 a^2 d^2\right )+\frac {b d (33 b c+2 a d)}{x}\right )}{15 b^2}+\sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x-\frac {\left (c^2 (b c+6 a d)\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{b}\\ &=-\frac {7}{5} d \sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^2-\frac {d \sqrt {a+\frac {b}{x}} \left (2 \left (57 b^2 c^2+15 a b c d-2 a^2 d^2\right )+\frac {b d (33 b c+2 a d)}{x}\right )}{15 b^2}+\sqrt {a+\frac {b}{x}} \left (c+\frac {d}{x}\right )^3 x+\frac {c^2 (b c+6 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 118, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+\frac {b}{x}} \left (4 a^2 d^3 x^2-2 a b d^2 x (d+15 c x)-3 b^2 \left (2 d^3+10 c d^2 x+30 c^2 d x^2-5 c^3 x^3\right )\right )}{15 b^2 x^2}+\frac {c^2 (b c+6 a d) \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b/x]*(4*a^2*d^3*x^2 - 2*a*b*d^2*x*(d + 15*c*x) - 3*b^2*(2*d^3 + 10*c*d^2*x + 30*c^2*d*x^2 - 5*c^3*x^
3)))/(15*b^2*x^2) + (c^2*(b*c + 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/Sqrt[a]

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Maple [A]
time = 0.07, size = 248, normalized size = 1.73

method	result	size
risch	\(\frac {\left (15 b^{2} c^{3} x^{3}+4 a^{2} d^{3} x^{2}-30 a c \,d^{2} x^{2} b -90 d \,c^{2} b^{2} x^{2}-2 a x \,d^{3} b -30 c \,d^{2} x \,b^{2}-6 b^{2} d^{3}\right ) \sqrt {\frac {a x +b}{x}}}{15 x^{2} b^{2}}+\frac {\left (3 c^{2} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {a}\, d +\frac {c^{3} \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) b}{2 \sqrt {a}}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{a x +b}\)	\(188\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, \left (180 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b \,c^{2} d \,x^{4}+30 \sqrt {a \,x^{2}+b x}\, \sqrt {a}\, b^{2} c^{3} x^{4}+90 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{2} c^{2} d \,x^{4}+15 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{3} c^{3} x^{4}-180 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b \,c^{2} d \,x^{2}+8 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} d^{3} x -60 d^{2} c \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b x \sqrt {a}-12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b \,d^{3}\right )}{30 x^{3} \sqrt {x \left (a x +b \right )}\, b^{2} \sqrt {a}}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d/x)^3*(a+1/x*b)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/30*((a*x+b)/x)^(1/2)/x^3*(180*(a*x^2+b*x)^(1/2)*a^(3/2)*b*c^2*d*x^4+30*(a*x^2+b*x)^(1/2)*a^(1/2)*b^2*c^3*x^4
+90*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*b^2*c^2*d*x^4+15*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/
2)+2*a*x+b)/a^(1/2))*b^3*c^3*x^4-180*(a*x^2+b*x)^(3/2)*a^(1/2)*b*c^2*d*x^2+8*(a*x^2+b*x)^(3/2)*a^(3/2)*d^3*x-6
0*d^2*c*(a*x^2+b*x)^(3/2)*b*x*a^(1/2)-12*(a*x^2+b*x)^(3/2)*a^(1/2)*b*d^3)/(x*(a*x+b))^(1/2)/b^2/a^(1/2)

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Maxima [A]
time = 0.49, size = 164, normalized size = 1.15 \begin {gather*} \frac {1}{2} \, {\left (2 \, \sqrt {a + \frac {b}{x}} x - \frac {b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{\sqrt {a}}\right )} c^{3} - 3 \, {\left (\sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) + 2 \, \sqrt {a + \frac {b}{x}}\right )} c^{2} d - \frac {2}{15} \, d^{3} {\left (\frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}}}{b^{2}} - \frac {5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a}{b^{2}}\right )} - \frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} c d^{2}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3*(a+b/x)^(1/2),x, algorithm="maxima")

[Out]

1/2*(2*sqrt(a + b/x)*x - b*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a))*c^3 - 3*(sqrt(a)*
log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) + 2*sqrt(a + b/x))*c^2*d - 2/15*d^3*(3*(a + b/x)^(5/2
)/b^2 - 5*(a + b/x)^(3/2)*a/b^2) - 2*(a + b/x)^(3/2)*c*d^2/b

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Fricas [A]
time = 3.24, size = 306, normalized size = 2.14 \begin {gather*} \left [\frac {15 \, {\left (b^{3} c^{3} + 6 \, a b^{2} c^{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 b^{2} c^{3} x^{3} - 6 \, a b^{2} d^{3} - 2 \, {\left (45 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} - 2 \, a^{3} d^{3}\right )} x^{2} - 2 \, {\left (15 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{30 \, a b^{2} x^{2}}, -\frac {15 \, {\left (b^{3} c^{3} + 6 \, a b^{2} c^{2} d\right )} \sqrt {-a} x^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (15 \, a b^{2} c^{3} x^{3} - 6 \, a b^{2} d^{3} - 2 \, {\left (45 \, a b^{2} c^{2} d + 15 \, a^{2} b c d^{2} - 2 \, a^{3} d^{3}\right )} x^{2} - 2 \, {\left (15 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{15 \, a b^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3*(a+b/x)^(1/2),x, algorithm="fricas")

[Out]

[1/30*(15*(b^3*c^3 + 6*a*b^2*c^2*d)*sqrt(a)*x^2*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(15*a*b^2*c
^3*x^3 - 6*a*b^2*d^3 - 2*(45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 2*a^3*d^3)*x^2 - 2*(15*a*b^2*c*d^2 + a^2*b*d^3)*x)
*sqrt((a*x + b)/x))/(a*b^2*x^2), -1/15*(15*(b^3*c^3 + 6*a*b^2*c^2*d)*sqrt(-a)*x^2*arctan(sqrt(-a)*sqrt((a*x +
b)/x)/a) - (15*a*b^2*c^3*x^3 - 6*a*b^2*d^3 - 2*(45*a*b^2*c^2*d + 15*a^2*b*c*d^2 - 2*a^3*d^3)*x^2 - 2*(15*a*b^2
*c*d^2 + a^2*b*d^3)*x)*sqrt((a*x + b)/x))/(a*b^2*x^2)]

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Sympy [A]
time = 30.67, size = 454, normalized size = 3.17 \begin {gather*} \frac {4 a^{\frac {11}{2}} b^{\frac {3}{2}} d^{3} 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 {5}{2}} d^{3} 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 {7}{2}} d^{3} 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 {9}{2}} d^{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 {4 a^{6} b d^{3} 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^{2} d^{3} 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 {6 a c^{2} d \operatorname {atan}{\left (\frac {\sqrt {a + \frac {b}{x}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + \sqrt {b} c^{3} \sqrt {x} \sqrt {\frac {a x}{b} + 1} - 6 c^{2} d \sqrt {a + \frac {b}{x}} + 3 c d^{2} \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 ) + \frac {b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)**3*(a+b/x)**(1/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3*(a+b/x)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

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Mupad [B]
time = 2.59, size = 173, normalized size = 1.21 \begin {gather*} {\left (a+\frac {b}{x}\right )}^{3/2}\,\left (\frac {6\,a\,d^3-6\,b\,c\,d^2}{3\,b^2}-\frac {4\,a\,d^3}{3\,b^2}\right )+\sqrt {a+\frac {b}{x}}\,\left (2\,a\,\left (\frac {6\,a\,d^3-6\,b\,c\,d^2}{b^2}-\frac {4\,a\,d^3}{b^2}\right )-\frac {6\,d\,{\left (a\,d-b\,c\right )}^2}{b^2}+\frac {2\,a^2\,d^3}{b^2}\right )+c^3\,x\,\sqrt {a+\frac {b}{x}}-\frac {2\,d^3\,{\left (a+\frac {b}{x}\right )}^{5/2}}{5\,b^2}-\frac {c^2\,\mathrm {atan}\left (\frac {\sqrt {a+\frac {b}{x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\left (6\,a\,d+b\,c\right )\,1{}\mathrm {i}}{\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(1/2)*(c + d/x)^3,x)

[Out]

(a + b/x)^(3/2)*((6*a*d^3 - 6*b*c*d^2)/(3*b^2) - (4*a*d^3)/(3*b^2)) + (a + b/x)^(1/2)*(2*a*((6*a*d^3 - 6*b*c*d
^2)/b^2 - (4*a*d^3)/b^2) - (6*d*(a*d - b*c)^2)/b^2 + (2*a^2*d^3)/b^2) + c^3*x*(a + b/x)^(1/2) - (2*d^3*(a + b/
x)^(5/2))/(5*b^2) - (c^2*atan(((a + b/x)^(1/2)*1i)/a^(1/2))*(6*a*d + b*c)*1i)/a^(1/2)

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