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

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

[Out]

c*(c+d/x)^2*x/a/(a+b/x)^(3/2)+1/3*(-a*d+b*c)*(15*b^3*c^2-4*a^3*d^2*x-2*a^2*b*d*(5*c*x+3*d)+a*b^2*c*(20*c*x-3*d
))/a^3/b^2/(a+b/x)^(3/2)/x-c^2*(-6*a*d+5*b*c)*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)

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Rubi [A]  time = 0.15, antiderivative size = 143, normalized size of antiderivative = 1.00, 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.43, size = 145, normalized size = 1.01 \[ \frac {\frac {4 a^5 d^3 x}{b^2}+\frac {6 a^4 d^2 (c x+d)}{b}+3 a^3 c^2 x (c x-8 d)+2 a^2 b c^2 (10 c x-9 d)+15 a b^2 c^3+3 a c^2 \sqrt {\frac {b}{a x}+1} (a x+b) (6 a d-5 b c) \tanh ^{-1}\left (\sqrt {\frac {b}{a x}+1}\right )}{3 a^4 \sqrt {a+\frac {b}{x}} (a x+b)} \]

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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fricas [A]  time = 0.98, size = 483, normalized size = 3.38 \[ \left [-\frac {3 \, {\left (5 \, b^{5} c^{3} - 6 \, a b^{4} c^{2} d + {\left (5 \, a^{2} b^{3} c^{3} - 6 \, a^{3} b^{2} c^{2} d\right )} x^{2} + 2 \, {\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d\right )} x\right )} \sqrt {a} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) - 2 \, {\left (3 \, a^{3} b^{2} c^{3} x^{3} + 2 \, {\left (10 \, a^{2} b^{3} c^{3} - 12 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{2} + 3 \, {\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d + 2 \, a^{4} b d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{6 \, {\left (a^{6} b^{2} x^{2} + 2 \, a^{5} b^{3} x + a^{4} b^{4}\right )}}, \frac {3 \, {\left (5 \, b^{5} c^{3} - 6 \, a b^{4} c^{2} d + {\left (5 \, a^{2} b^{3} c^{3} - 6 \, a^{3} b^{2} c^{2} d\right )} x^{2} + 2 \, {\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (3 \, a^{3} b^{2} c^{3} x^{3} + 2 \, {\left (10 \, a^{2} b^{3} c^{3} - 12 \, a^{3} b^{2} c^{2} d + 3 \, a^{4} b c d^{2} + 2 \, a^{5} d^{3}\right )} x^{2} + 3 \, {\left (5 \, a b^{4} c^{3} - 6 \, a^{2} b^{3} c^{2} d + 2 \, a^{4} b d^{3}\right )} x\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{6} b^{2} x^{2} + 2 \, a^{5} b^{3} x + a^{4} b^{4}\right )}}\right ] \]

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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giac [A]  time = 0.26, size = 203, normalized size = 1.42 \[ -\frac {\frac {3 \, b^{2} c^{3} \sqrt {\frac {a x + b}{x}}}{{\left (a - \frac {a x + b}{x}\right )} a^{3}} - \frac {3 \, {\left (5 \, b^{2} c^{3} - 6 \, a b c^{2} d\right )} \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} - \frac {2 \, {\left (a b^{3} c^{3} - 3 \, a^{2} b^{2} c^{2} d + 3 \, a^{3} b c d^{2} - a^{4} d^{3} + \frac {6 \, {\left (a x + b\right )} b^{3} c^{3}}{x} - \frac {9 \, {\left (a x + b\right )} a b^{2} c^{2} d}{x} + \frac {3 \, {\left (a x + b\right )} a^{3} d^{3}}{x}\right )} x}{{\left (a x + b\right )} a^{3} b \sqrt {\frac {a x + b}{x}}}}{3 \, b} \]

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*(3*b^2*c^3*sqrt((a*x + b)/x)/((a - (a*x + b)/x)*a^3) - 3*(5*b^2*c^3 - 6*a*b*c^2*d)*arctan(sqrt((a*x + b)/
x)/sqrt(-a))/(sqrt(-a)*a^3) - 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*sqrt((a*x + b)/x)))/b

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maple [B]  time = 0.06, size = 1150, normalized size = 8.04 \[ \text {result too large to display} \]

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

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maxima [A]  time = 1.34, size = 228, normalized size = 1.59 \[ \frac {1}{6} \, c^{3} {\left (\frac {2 \, {\left (15 \, {\left (a + \frac {b}{x}\right )}^{2} b - 10 \, {\left (a + \frac {b}{x}\right )} a b - 2 \, a^{2} b\right )}}{{\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{3} - {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{4}} + \frac {15 \, b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )} - c^{2} d {\left (\frac {3 \, \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (4 \, a + \frac {3 \, b}{x}\right )}}{{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2}}\right )} + \frac {2}{3} \, d^{3} {\left (\frac {3}{\sqrt {a + \frac {b}{x}} b^{2}} - \frac {a}{{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {2 \, c d^{2}}{{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b} \]

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]

1/6*c^3*(2*(15*(a + b/x)^2*b - 10*(a + b/x)*a*b - 2*a^2*b)/((a + b/x)^(5/2)*a^3 - (a + b/x)^(3/2)*a^4) + 15*b*
log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2)) - c^2*d*(3*log((sqrt(a + b/x) - sqrt(a))/(sq
rt(a + b/x) + sqrt(a)))/a^(5/2) + 2*(4*a + 3*b/x)/((a + b/x)^(3/2)*a^2)) + 2/3*d^3*(3/(sqrt(a + b/x)*b^2) - a/
((a + b/x)^(3/2)*b^2)) + 2*c*d^2/((a + b/x)^(3/2)*b)

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mupad [B]  time = 2.05, size = 194, normalized size = 1.36 \[ \frac {\frac {2\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{3\,a}+\frac {{\left (a+\frac {b}{x}\right )}^2\,\left (2\,a^3\,d^3-6\,a\,b^2\,c^2\,d+5\,b^3\,c^3\right )}{a^3}-\frac {2\,\left (a+\frac {b}{x}\right )\,\left (4\,a^3\,d^3-3\,a^2\,b\,c\,d^2-6\,a\,b^2\,c^2\,d+5\,b^3\,c^3\right )}{3\,a^2}}{b^2\,{\left (a+\frac {b}{x}\right )}^{5/2}-a\,b^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}+\frac {c^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )\,\left (6\,a\,d-5\,b\,c\right )}{a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d/x)^3/(a + b/x)^(5/2),x)

[Out]

((2*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(3*a) + ((a + b/x)^2*(2*a^3*d^3 + 5*b^3*c^3 - 6*a*b^2
*c^2*d))/a^3 - (2*(a + b/x)*(4*a^3*d^3 + 5*b^3*c^3 - 6*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(3*a^2))/(b^2*(a + b/x)^(
5/2) - a*b^2*(a + b/x)^(3/2)) + (c^2*atanh((a + b/x)^(1/2)/a^(1/2))*(6*a*d - 5*b*c))/a^(7/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x + d\right )^{3}}{x^{3} \left (a + \frac {b}{x}\right )^{\frac {5}{2}}}\, dx \]

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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