3.252 \(\int \frac{(c+\frac{d}{x})^3}{(a+\frac{b}{x})^{3/2}} \, dx\)

Optimal. Leaf size=132 \[ \frac{(b c-2 a d) \left (2 a^2 d^2-2 a b c d+3 b^2 c^2\right )-\frac{a b d^2 (2 a d+b c)}{x}}{a^2 b^2 \sqrt{a+\frac{b}{x}}}-\frac{3 c^2 (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{c x \left (c+\frac{d}{x}\right )^2}{a \sqrt{a+\frac{b}{x}}} \]

[Out]

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

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Rubi [A]  time = 0.10126, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {375, 98, 146, 63, 208} \[ \frac{(b c-2 a d) \left (2 a^2 d^2-2 a b c d+3 b^2 c^2\right )-\frac{a b d^2 (2 a d+b c)}{x}}{a^2 b^2 \sqrt{a+\frac{b}{x}}}-\frac{3 c^2 (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2}}+\frac{c x \left (c+\frac{d}{x}\right )^2}{a \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d/x)^3/(a + b/x)^(3/2),x]

[Out]

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

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 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(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*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && 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 \frac{\left (c+\frac{d}{x}\right )^3}{\left (a+\frac{b}{x}\right )^{3/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{(c+d x)^3}{x^2 (a+b x)^{3/2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \sqrt{a+\frac{b}{x}}}+\frac{\operatorname{Subst}\left (\int \frac{(c+d x) \left (\frac{3}{2} c (b c-2 a d)-\frac{1}{2} d (b c+2 a d) x\right )}{x (a+b x)^{3/2}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{(b c-2 a d) \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right )-\frac{a b d^2 (b c+2 a d)}{x}}{a^2 b^2 \sqrt{a+\frac{b}{x}}}+\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \sqrt{a+\frac{b}{x}}}+\frac{\left (3 c^2 (b c-2 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 a^2}\\ &=\frac{(b c-2 a d) \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right )-\frac{a b d^2 (b c+2 a d)}{x}}{a^2 b^2 \sqrt{a+\frac{b}{x}}}+\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \sqrt{a+\frac{b}{x}}}+\frac{\left (3 c^2 (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 )}{a^2 b}\\ &=\frac{(b c-2 a d) \left (3 b^2 c^2-2 a b c d+2 a^2 d^2\right )-\frac{a b d^2 (b c+2 a d)}{x}}{a^2 b^2 \sqrt{a+\frac{b}{x}}}+\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \sqrt{a+\frac{b}{x}}}-\frac{3 c^2 (b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0546606, size = 92, normalized size = 0.7 \[ \frac{a \left (-4 a^2 d^3 x-2 a b d^2 (d-3 c x)+b^2 c^3 x^2\right )+3 b^2 c^2 x (b c-2 a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b}{a x}+1\right )}{a^2 b^2 x \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d/x)^3/(a + b/x)^(3/2),x]

[Out]

(a*(-4*a^2*d^3*x + b^2*c^3*x^2 - 2*a*b*d^2*(d - 3*c*x)) + 3*b^2*c^2*(b*c - 2*a*d)*x*Hypergeometric2F1[-1/2, 1,
1/2, 1 + b/(a*x)])/(a^2*b^2*Sqrt[a + b/x]*x)

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Maple [B]  time = 0.016, size = 969, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d/x)^3/(a+b/x)^(3/2),x)

[Out]

-1/2*((a*x+b)/x)^(1/2)/x/a^(5/2)*(3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^4*a^4*b^2*c*d^2+3*
ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*b^6*c^3+4*(a*x^2+b*x)^(3/2)*a^(9/2)*x^2*d^3-4*a^(9/2
)*((a*x+b)*x)^(3/2)*x^2*d^3+4*(a*x^2+b*x)^(3/2)*a^(5/2)*b^2*d^3+4*a^(3/2)*((a*x+b)*x)^(3/2)*x^2*b^3*c^3-12*a^(
3/2)*((a*x+b)*x)^(1/2)*x^3*b^4*c^3+8*(a*x^2+b*x)^(3/2)*a^(7/2)*x*b*d^3-6*a^(5/2)*((a*x+b)*x)^(1/2)*x^4*b^3*c^3
+3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^4*a^2*b^4*c^3+6*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)
+2*a*x+b)/a^(1/2))*x^3*a*b^5*c^3-6*a^(1/2)*((a*x+b)*x)^(1/2)*x^2*b^5*c^3-3*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)
+2*a*x+b)/a^(1/2))*x^2*a^2*b^4*c*d^2+3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^2*b^4*c*d^2
-6*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a*b^5*c^2*d-3*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)
+2*a*x+b)/a^(1/2))*x^4*a^4*b^2*c*d^2-6*(a*x^2+b*x)^(1/2)*a^(9/2)*x^4*b*c*d^2-6*a^(9/2)*((a*x+b)*x)^(1/2)*x^4*b
*c*d^2+12*a^(7/2)*((a*x+b)*x)^(1/2)*x^4*b^2*c^2*d-12*(a*x^2+b*x)^(1/2)*a^(7/2)*x^3*b^2*c*d^2+12*a^(7/2)*((a*x+
b)*x)^(3/2)*x^2*b*c*d^2-12*a^(5/2)*((a*x+b)*x)^(3/2)*x^2*b^2*c^2*d-12*a^(7/2)*((a*x+b)*x)^(1/2)*x^3*b^2*c*d^2+
24*a^(5/2)*((a*x+b)*x)^(1/2)*x^3*b^3*c^2*d-6*(a*x^2+b*x)^(1/2)*a^(5/2)*x^2*b^3*c*d^2-6*a^(5/2)*((a*x+b)*x)^(1/
2)*x^2*b^3*c*d^2+12*a^(3/2)*((a*x+b)*x)^(1/2)*x^2*b^4*c^2*d-6*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(
1/2))*x^4*a^3*b^3*c^2*d-6*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^3*a^3*b^3*c*d^2+6*ln(1/2*(2*
((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^3*a^3*b^3*c*d^2-12*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/
a^(1/2))*x^3*a^2*b^4*c^2*d)/((a*x+b)*x)^(1/2)/b^3/(a*x+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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32398, size = 699, normalized size = 5.3 \begin{align*} \left [-\frac{3 \,{\left (b^{4} c^{3} - 2 \, a b^{3} c^{2} d +{\left (a b^{3} c^{3} - 2 \, a^{2} b^{2} c^{2} d\right )} x\right )} \sqrt{a} \log \left (2 \, a x + 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) - 2 \,{\left (a^{2} b^{2} c^{3} x^{2} - 2 \, a^{3} b d^{3} +{\left (3 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 6 \, a^{3} b c d^{2} - 4 \, a^{4} d^{3}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{2 \,{\left (a^{4} b^{2} x + a^{3} b^{3}\right )}}, \frac{3 \,{\left (b^{4} c^{3} - 2 \, a b^{3} c^{2} d +{\left (a b^{3} c^{3} - 2 \, a^{2} b^{2} c^{2} d\right )} x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (a^{2} b^{2} c^{3} x^{2} - 2 \, a^{3} b d^{3} +{\left (3 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 6 \, a^{3} b c d^{2} - 4 \, a^{4} d^{3}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{a^{4} b^{2} x + a^{3} b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(3*(b^4*c^3 - 2*a*b^3*c^2*d + (a*b^3*c^3 - 2*a^2*b^2*c^2*d)*x)*sqrt(a)*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x
+ b)/x) + b) - 2*(a^2*b^2*c^3*x^2 - 2*a^3*b*d^3 + (3*a*b^3*c^3 - 6*a^2*b^2*c^2*d + 6*a^3*b*c*d^2 - 4*a^4*d^3)
*x)*sqrt((a*x + b)/x))/(a^4*b^2*x + a^3*b^3), (3*(b^4*c^3 - 2*a*b^3*c^2*d + (a*b^3*c^3 - 2*a^2*b^2*c^2*d)*x)*s
qrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (a^2*b^2*c^3*x^2 - 2*a^3*b*d^3 + (3*a*b^3*c^3 - 6*a^2*b^2*c^2*d
+ 6*a^3*b*c*d^2 - 4*a^4*d^3)*x)*sqrt((a*x + b)/x))/(a^4*b^2*x + a^3*b^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x + d\right )^{3}}{x^{3} \left (a + \frac{b}{x}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20549, size = 297, normalized size = 2.25 \begin{align*} -b{\left (\frac{2 \, d^{3} \sqrt{\frac{a x + b}{x}}}{b^{3}} - \frac{3 \,{\left (b c^{3} - 2 \, a c^{2} d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2} b} - \frac{2 \, a b^{3} c^{3} - 6 \, a^{2} b^{2} c^{2} d + 6 \, a^{3} b c d^{2} - 2 \, a^{4} d^{3} - \frac{3 \,{\left (a x + b\right )} b^{3} c^{3}}{x} + \frac{6 \,{\left (a x + b\right )} a b^{2} c^{2} d}{x} - \frac{6 \,{\left (a x + b\right )} a^{2} b c d^{2}}{x} + \frac{2 \,{\left (a x + b\right )} a^{3} d^{3}}{x}}{{\left (a \sqrt{\frac{a x + b}{x}} - \frac{{\left (a x + b\right )} \sqrt{\frac{a x + b}{x}}}{x}\right )} a^{2} b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*(2*d^3*sqrt((a*x + b)/x)/b^3 - 3*(b*c^3 - 2*a*c^2*d)*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a^2*b) -
(2*a*b^3*c^3 - 6*a^2*b^2*c^2*d + 6*a^3*b*c*d^2 - 2*a^4*d^3 - 3*(a*x + b)*b^3*c^3/x + 6*(a*x + b)*a*b^2*c^2*d/x
- 6*(a*x + b)*a^2*b*c*d^2/x + 2*(a*x + b)*a^3*d^3/x)/((a*sqrt((a*x + b)/x) - (a*x + b)*sqrt((a*x + b)/x)/x)*a
^2*b^3))