3.259 \(\int \frac{(c+\frac{d}{x})^3}{(a+\frac{b}{x})^{5/2}} \, dx\)

Optimal. Leaf size=143 \[ \frac{(b c-a d) \left (-2 a^2 b d (5 c x+3 d)-4 a^3 d^2 x+a b^2 c (20 c x-3 d)+15 b^3 c^2\right )}{3 a^3 b^2 x \left (a+\frac{b}{x}\right )^{3/2}}-\frac{c^2 (5 b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{c x \left (c+\frac{d}{x}\right )^2}{a \left (a+\frac{b}{x}\right )^{3/2}} \]

[Out]

(c*(c + d/x)^2*x)/(a*(a + b/x)^(3/2)) + ((b*c - a*d)*(15*b^3*c^2 - 4*a^3*d^2*x - 2*a^2*b*d*(3*d + 5*c*x) + a*b
^2*c*(-3*d + 20*c*x)))/(3*a^3*b^2*(a + b/x)^(3/2)*x) - (c^2*(5*b*c - 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/a^
(7/2)

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Rubi [A]  time = 0.151227, antiderivative size = 143, 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, 145, 63, 208} \[ \frac{(b c-a d) \left (-2 a^2 b d (5 c x+3 d)-4 a^3 d^2 x+a b^2 c (20 c x-3 d)+15 b^3 c^2\right )}{3 a^3 b^2 x \left (a+\frac{b}{x}\right )^{3/2}}-\frac{c^2 (5 b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{c x \left (c+\frac{d}{x}\right )^2}{a \left (a+\frac{b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d/x)^3/(a + b/x)^(5/2),x]

[Out]

(c*(c + d/x)^2*x)/(a*(a + b/x)^(3/2)) + ((b*c - a*d)*(15*b^3*c^2 - 4*a^3*d^2*x - 2*a^2*b*d*(3*d + 5*c*x) + a*b
^2*c*(-3*d + 20*c*x)))/(3*a^3*b^2*(a + b/x)^(3/2)*x) - (c^2*(5*b*c - 6*a*d)*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/a^
(7/2)

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 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

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 )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{(c+d x)^3}{x^2 (a+b x)^{5/2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \left (a+\frac{b}{x}\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{(c+d x) \left (\frac{1}{2} c (5 b c-6 a d)+\frac{1}{2} d (b c-2 a d) x\right )}{x (a+b x)^{5/2}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \left (a+\frac{b}{x}\right )^{3/2}}+\frac{(b c-a d) \left (15 b^3 c^2-4 a^3 d^2 x-a b^2 c (3 d-20 c x)-2 a^2 b d (3 d+5 c x)\right )}{3 a^3 b^2 \left (a+\frac{b}{x}\right )^{3/2} x}+\frac{\left (c^2 (5 b c-6 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{2 a^3}\\ &=\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \left (a+\frac{b}{x}\right )^{3/2}}+\frac{(b c-a d) \left (15 b^3 c^2-4 a^3 d^2 x-a b^2 c (3 d-20 c x)-2 a^2 b d (3 d+5 c x)\right )}{3 a^3 b^2 \left (a+\frac{b}{x}\right )^{3/2} x}+\frac{\left (c^2 (5 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 )}{a^3 b}\\ &=\frac{c \left (c+\frac{d}{x}\right )^2 x}{a \left (a+\frac{b}{x}\right )^{3/2}}+\frac{(b c-a d) \left (15 b^3 c^2-4 a^3 d^2 x-a b^2 c (3 d-20 c x)-2 a^2 b d (3 d+5 c x)\right )}{3 a^3 b^2 \left (a+\frac{b}{x}\right )^{3/2} x}-\frac{c^2 (5 b c-6 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.294236, size = 145, normalized size = 1.01 \[ \frac{\frac{4 a^5 d^3 x}{b^2}+2 a^2 b c^2 (10 c x-9 d)+\frac{6 a^4 d^2 (c x+d)}{b}+3 a^3 c^2 x (c x-8 d)+15 a b^2 c^3+3 a c^2 \sqrt{\frac{b}{a x}+1} (a x+b) (6 a d-5 b c) \tanh ^{-1}\left (\sqrt{\frac{b}{a x}+1}\right )}{3 a^4 \sqrt{a+\frac{b}{x}} (a x+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d/x)^3/(a + b/x)^(5/2),x]

[Out]

(15*a*b^2*c^3 + (4*a^5*d^3*x)/b^2 + 3*a^3*c^2*x*(-8*d + c*x) + (6*a^4*d^2*(d + c*x))/b + 2*a^2*b*c^2*(-9*d + 1
0*c*x) + 3*a*c^2*(-5*b*c + 6*a*d)*Sqrt[1 + b/(a*x)]*(b + a*x)*ArcTanh[Sqrt[1 + b/(a*x)]])/(3*a^4*Sqrt[a + b/x]
*(b + a*x))

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Maple [B]  time = 0.014, size = 1150, normalized size = 8. \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)^(5/2),x)

[Out]

1/6*((a*x+b)/x)^(1/2)*x/a^(7/2)*(3*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^3*b^4*d^3+30*a^(1/2
)*((a*x+b)*x)^(1/2)*b^6*c^3-3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a^3*b^4*d^3-16*a^(9/2)*((a
*x+b)*x)^(3/2)*b*d^3-20*a^(3/2)*((a*x+b)*x)^(3/2)*b^4*c^3+6*(a*x^2+b*x)^(1/2)*a^(7/2)*b^3*d^3+6*a^(7/2)*((a*x+
b)*x)^(1/2)*b^3*d^3+6*(a*x^2+b*x)^(1/2)*a^(13/2)*x^3*d^3+6*a^(13/2)*((a*x+b)*x)^(1/2)*x^3*d^3-12*a^(11/2)*((a*
x+b)*x)^(3/2)*x*d^3-15*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*b^7*c^3+18*a^(9/2)*((a*x+b)*x)^(1
/2)*x*b^2*d^3+90*a^(3/2)*((a*x+b)*x)^(1/2)*x*b^5*c^3-36*a^(3/2)*((a*x+b)*x)^(1/2)*b^5*c^2*d+18*ln(1/2*(2*((a*x
+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*b^6*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^3
*a^6*b*d^3-3*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^3*a^6*b*d^3-15*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^4*c^3+9*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^5*b^
2*d^3-9*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^5*b^2*d^3-45*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^5*c^3+90*a^(5/2)*((a*x+b)*x)^(1/2)*x^2*b^4*c^3+18*(a*x^2+b*x)^(1/2)*a^(9/2)
*x*b^2*d^3+12*a^(7/2)*((a*x+b)*x)^(3/2)*b^2*c*d^2+24*a^(5/2)*((a*x+b)*x)^(3/2)*b^3*c^2*d-9*ln(1/2*(2*((a*x+b)*
x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^4*b^3*d^3-45*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a*
b^6*c^3+30*a^(7/2)*((a*x+b)*x)^(1/2)*x^3*b^3*c^3+9*ln(1/2*(2*(a*x^2+b*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^4
*b^3*d^3+18*(a*x^2+b*x)^(1/2)*a^(11/2)*x^2*b*d^3-24*a^(5/2)*((a*x+b)*x)^(3/2)*x*b^3*c^3+18*a^(11/2)*((a*x+b)*x
)^(1/2)*x^2*b*d^3+18*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^3*a^4*b^3*c^2*d-108*a^(7/2)*((a*x
+b)*x)^(1/2)*x^2*b^3*c^2*d-108*a^(5/2)*((a*x+b)*x)^(1/2)*x*b^4*c^2*d-36*a^(9/2)*((a*x+b)*x)^(1/2)*x^3*b^2*c^2*
d+36*a^(7/2)*((a*x+b)*x)^(3/2)*x*b^2*c^2*d+54*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^3*b^
4*c^2*d+54*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^2*b^5*c^2*d)/((a*x+b)*x)^(1/2)/b^3/(a*x+b
)^3

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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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32662, size = 1002, normalized size = 7.01 \begin{align*} \left [-\frac{3 \,{\left (5 \, b^{5} c^{3} - 6 \, a b^{4} c^{2} d +{\left (5 \, a^{2} b^{3} c^{3} - 6 \, a^{3} b^{2} c^{2} d\right )} x^{2} + 2 \,{\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} 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 (3 \, a^{3} b^{2} c^{3} x^{3} + 2 \,{\left (10 \, a^{2} b^{3} c^{3} - 12 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{2} + 3 \,{\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d + 2 \, a^{4} b d^{3}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{6 \,{\left (a^{6} b^{2} x^{2} + 2 \, a^{5} b^{3} x + a^{4} b^{4}\right )}}, \frac{3 \,{\left (5 \, b^{5} c^{3} - 6 \, a b^{4} c^{2} d +{\left (5 \, a^{2} b^{3} c^{3} - 6 \, a^{3} b^{2} c^{2} d\right )} x^{2} + 2 \,{\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d\right )} x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (3 \, a^{3} b^{2} c^{3} x^{3} + 2 \,{\left (10 \, a^{2} b^{3} c^{3} - 12 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{2} + 3 \,{\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d + 2 \, a^{4} b d^{3}\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{3 \,{\left (a^{6} b^{2} x^{2} + 2 \, a^{5} b^{3} x + a^{4} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3/(a+b/x)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(5*b^5*c^3 - 6*a*b^4*c^2*d + (5*a^2*b^3*c^3 - 6*a^3*b^2*c^2*d)*x^2 + 2*(5*a*b^4*c^3 - 6*a^2*b^3*c^2*d
)*x)*sqrt(a)*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(3*a^3*b^2*c^3*x^3 + 2*(10*a^2*b^3*c^3 - 12*a^
3*b^2*c^2*d + 3*a^4*b*c*d^2 + 2*a^5*d^3)*x^2 + 3*(5*a*b^4*c^3 - 6*a^2*b^3*c^2*d + 2*a^4*b*d^3)*x)*sqrt((a*x +
b)/x))/(a^6*b^2*x^2 + 2*a^5*b^3*x + a^4*b^4), 1/3*(3*(5*b^5*c^3 - 6*a*b^4*c^2*d + (5*a^2*b^3*c^3 - 6*a^3*b^2*c
^2*d)*x^2 + 2*(5*a*b^4*c^3 - 6*a^2*b^3*c^2*d)*x)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3*a^3*b^2*c^
3*x^3 + 2*(10*a^2*b^3*c^3 - 12*a^3*b^2*c^2*d + 3*a^4*b*c*d^2 + 2*a^5*d^3)*x^2 + 3*(5*a*b^4*c^3 - 6*a^2*b^3*c^2
*d + 2*a^4*b*d^3)*x)*sqrt((a*x + b)/x))/(a^6*b^2*x^2 + 2*a^5*b^3*x + a^4*b^4)]

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)**3/(a+b/x)**(5/2),x)

[Out]

Integral((c*x + d)**3/(x**3*(a + b/x)**(5/2)), x)

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Giac [A]  time = 1.19536, size = 267, normalized size = 1.87 \begin{align*} -\frac{1}{3} \, b{\left (\frac{3 \, c^{3} \sqrt{\frac{a x + b}{x}}}{{\left (a - \frac{a x + b}{x}\right )} a^{3}} - \frac{3 \,{\left (5 \, b c^{3} - 6 \, a c^{2} d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3} b} - \frac{2 \,{\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3} + \frac{6 \,{\left (a x + b\right )} b^{3} c^{3}}{x} - \frac{9 \,{\left (a x + b\right )} a b^{2} c^{2} d}{x} + \frac{3 \,{\left (a x + b\right )} a^{3} d^{3}}{x}\right )} x}{{\left (a x + b\right )} a^{3} b^{3} \sqrt{\frac{a x + b}{x}}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d/x)^3/(a+b/x)^(5/2),x, algorithm="giac")

[Out]

-1/3*b*(3*c^3*sqrt((a*x + b)/x)/((a - (a*x + b)/x)*a^3) - 3*(5*b*c^3 - 6*a*c^2*d)*arctan(sqrt((a*x + b)/x)/sqr
t(-a))/(sqrt(-a)*a^3*b) - 2*(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3 + 6*(a*x + b)*b^3*c^3/x - 9
*(a*x + b)*a*b^2*c^2*d/x + 3*(a*x + b)*a^3*d^3/x)*x/((a*x + b)*a^3*b^3*sqrt((a*x + b)/x)))