3.224 \(\int \sqrt{a+\frac{b}{x}} (c+\frac{d}{x})^3 \, dx\)

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

[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]

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Rubi [A]  time = 0.123315, antiderivative size = 143, normalized size of antiderivative = 1., 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 = {375, 97, 153, 147, 63, 208} \[ -\frac{d \sqrt{a+\frac{b}{x}} \left (2 \left (-2 a^2 d^2+15 a b c d+57 b^2 c^2\right )+\frac{b d (2 a d+33 b c)}{x}\right )}{15 b^2}+\frac{c^2 (6 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}}+x \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )^3-\frac{7}{5} d \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )^2 \]

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 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 97

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 153

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 147

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 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 \sqrt{a+\frac{b}{x}} \left (c+\frac{d}{x}\right )^3 \, dx &=-\operatorname{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-\operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.145502, size = 118, normalized size = 0.83 \[ \frac{\sqrt{a+\frac{b}{x}} \left (4 a^2 d^3 x^2-2 a b d^2 x (15 c x+d)-3 b^2 \left (30 c^2 d x^2-5 c^3 x^3+10 c d^2 x+2 d^3\right )\right )}{15 b^2 x^2}+\frac{c^2 (6 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{\sqrt{a}} \]

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 = 0.016, size = 248, normalized size = 1.7 \begin{align*}{\frac{1}{30\,{b}^{2}{x}^{3}}\sqrt{{\frac{ax+b}{x}}} \left ( 90\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{4}a{b}^{2}{c}^{2}d+15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{4}{b}^{3}{c}^{3}+180\,{a}^{3/2}\sqrt{a{x}^{2}+bx}{x}^{4}b{c}^{2}d+30\,\sqrt{a}\sqrt{a{x}^{2}+bx}{x}^{4}{b}^{2}{c}^{3}-180\,\sqrt{a} \left ( a{x}^{2}+bx \right ) ^{3/2}{x}^{2}b{c}^{2}d+8\,{a}^{3/2} \left ( a{x}^{2}+bx \right ) ^{3/2}x{d}^{3}-60\,{d}^{2}c \left ( a{x}^{2}+bx \right ) ^{3/2}\sqrt{a}xb-12\,\sqrt{a} \left ( a{x}^{2}+bx \right ) ^{3/2}b{d}^{3} \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((c+d/x)^3*(a+b/x)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.27133, size = 672, normalized size = 4.7 \begin{align*} \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{align*}

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 = 26.27, size = 454, normalized size = 3.17 \begin{align*} \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{align*}

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

[Out]

Exception raised: TypeError