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

Optimal. Leaf size=122 \[ \frac{2 a^2 d^2+b c (5 b c-4 a d)}{3 a^2 b \left (a+\frac{b}{x}\right )^{3/2}}+\frac{c (5 b c-4 a d)}{a^3 \sqrt{a+\frac{b}{x}}}-\frac{c (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{c^2 x}{a \left (a+\frac{b}{x}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0951476, antiderivative size = 118, normalized size of antiderivative = 0.97, 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, 89, 78, 51, 63, 208} \[ \frac{\frac{c (5 b c-4 a d)}{a^2}+\frac{2 d^2}{b}}{3 \left (a+\frac{b}{x}\right )^{3/2}}+\frac{c (5 b c-4 a d)}{a^3 \sqrt{a+\frac{b}{x}}}-\frac{c (5 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{a^{7/2}}+\frac{c^2 x}{a \left (a+\frac{b}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d/x)^2/(a + b/x)^(5/2),x]

[Out]

((2*d^2)/b + (c*(5*b*c - 4*a*d))/a^2)/(3*(a + b/x)^(3/2)) + (c*(5*b*c - 4*a*d))/(a^3*Sqrt[a + b/x]) + (c^2*x)/
(a*(a + b/x)^(3/2)) - (c*(5*b*c - 4*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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [C]  time = 0.0491668, size = 97, normalized size = 0.8 \[ \frac{a x \left (2 a^2 d^2+a b c (3 c x-4 d)+5 b^2 c^2\right )+3 b c (a x+b) (5 b c-4 a d) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b}{a x}+1\right )}{3 a^3 b \sqrt{a+\frac{b}{x}} (a x+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(5*b^2*c^2 + 2*a^2*d^2 + a*b*c*(-4*d + 3*c*x)) + 3*b*c*(5*b*c - 4*a*d)*(b + a*x)*Hypergeometric2F1[-1/2,
1, 1/2, 1 + b/(a*x)])/(3*a^3*b*Sqrt[a + b/x]*(b + a*x))

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Maple [B]  time = 0.011, size = 588, normalized size = 4.8 \begin{align*}{\frac{x}{6\,b \left ( ax+b \right ) ^{3}}\sqrt{{\frac{ax+b}{x}}} \left ( 12\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{3}{a}^{4}bcd-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{3}{a}^{3}{b}^{2}{c}^{2}-24\,{a}^{9/2}\sqrt{ \left ( ax+b \right ) x}{x}^{3}cd+30\,{a}^{7/2}\sqrt{ \left ( ax+b \right ) x}{x}^{3}b{c}^{2}+36\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{2}{a}^{3}{b}^{2}cd-45\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){x}^{2}{a}^{2}{b}^{3}{c}^{2}+24\,{a}^{7/2} \left ( \left ( ax+b \right ) x \right ) ^{3/2}xcd-24\,{a}^{5/2} \left ( \left ( ax+b \right ) x \right ) ^{3/2}xb{c}^{2}-72\,{a}^{7/2}\sqrt{ \left ( ax+b \right ) x}{x}^{2}bcd+90\,{a}^{5/2}\sqrt{ \left ( ax+b \right ) x}{x}^{2}{b}^{2}{c}^{2}+36\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) x{a}^{2}{b}^{3}cd-45\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) xa{b}^{4}{c}^{2}+4\,{a}^{7/2} \left ( \left ( ax+b \right ) x \right ) ^{3/2}{d}^{2}+16\,{a}^{5/2} \left ( \left ( ax+b \right ) x \right ) ^{3/2}bcd-20\,{a}^{3/2} \left ( \left ( ax+b \right ) x \right ) ^{3/2}{b}^{2}{c}^{2}-72\,{a}^{5/2}\sqrt{ \left ( ax+b \right ) x}x{b}^{2}cd+90\,{a}^{3/2}\sqrt{ \left ( ax+b \right ) x}x{b}^{3}{c}^{2}+12\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) a{b}^{4}cd-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{ \left ( ax+b \right ) x}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ){b}^{5}{c}^{2}-24\,{a}^{3/2}\sqrt{ \left ( ax+b \right ) x}{b}^{3}cd+30\,\sqrt{a}\sqrt{ \left ( ax+b \right ) x}{b}^{4}{c}^{2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d/x)^2/(a+b/x)^(5/2),x)

[Out]

1/6*((a*x+b)/x)^(1/2)*x/a^(7/2)*(12*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^3*a^4*b*c*d-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^2*c^2-24*a^(9/2)*((a*x+b)*x)^(1/2)*x^3*c*d+30*a^(
7/2)*((a*x+b)*x)^(1/2)*x^3*b*c^2+36*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^2*c*d-45*l
n(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x^2*a^2*b^3*c^2+24*a^(7/2)*((a*x+b)*x)^(3/2)*x*c*d-24*a^(
5/2)*((a*x+b)*x)^(3/2)*x*b*c^2-72*a^(7/2)*((a*x+b)*x)^(1/2)*x^2*b*c*d+90*a^(5/2)*((a*x+b)*x)^(1/2)*x^2*b^2*c^2
+36*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*x*a^2*b^3*c*d-45*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)
+2*a*x+b)/a^(1/2))*x*a*b^4*c^2+4*a^(7/2)*((a*x+b)*x)^(3/2)*d^2+16*a^(5/2)*((a*x+b)*x)^(3/2)*b*c*d-20*a^(3/2)*(
(a*x+b)*x)^(3/2)*b^2*c^2-72*a^(5/2)*((a*x+b)*x)^(1/2)*x*b^2*c*d+90*a^(3/2)*((a*x+b)*x)^(1/2)*x*b^3*c^2+12*ln(1
/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a*b^4*c*d-15*ln(1/2*(2*((a*x+b)*x)^(1/2)*a^(1/2)+2*a*x+b)/a^
(1/2))*b^5*c^2-24*a^(3/2)*((a*x+b)*x)^(1/2)*b^3*c*d+30*a^(1/2)*((a*x+b)*x)^(1/2)*b^4*c^2)/((a*x+b)*x)^(1/2)/b/
(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)^2/(a+b/x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36801, size = 864, normalized size = 7.08 \begin{align*} \left [-\frac{3 \,{\left (5 \, b^{4} c^{2} - 4 \, a b^{3} c d +{\left (5 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d\right )} x^{2} + 2 \,{\left (5 \, a b^{3} c^{2} - 4 \, a^{2} b^{2} c 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 c^{2} x^{3} + 2 \,{\left (10 \, a^{2} b^{2} c^{2} - 8 \, a^{3} b c d + a^{4} d^{2}\right )} x^{2} + 3 \,{\left (5 \, a b^{3} c^{2} - 4 \, a^{2} b^{2} c d\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{6 \,{\left (a^{6} b x^{2} + 2 \, a^{5} b^{2} x + a^{4} b^{3}\right )}}, \frac{3 \,{\left (5 \, b^{4} c^{2} - 4 \, a b^{3} c d +{\left (5 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d\right )} x^{2} + 2 \,{\left (5 \, a b^{3} c^{2} - 4 \, a^{2} b^{2} c d\right )} x\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (3 \, a^{3} b c^{2} x^{3} + 2 \,{\left (10 \, a^{2} b^{2} c^{2} - 8 \, a^{3} b c d + a^{4} d^{2}\right )} x^{2} + 3 \,{\left (5 \, a b^{3} c^{2} - 4 \, a^{2} b^{2} c d\right )} x\right )} \sqrt{\frac{a x + b}{x}}}{3 \,{\left (a^{6} b x^{2} + 2 \, a^{5} b^{2} x + a^{4} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*(5*b^4*c^2 - 4*a*b^3*c*d + (5*a^2*b^2*c^2 - 4*a^3*b*c*d)*x^2 + 2*(5*a*b^3*c^2 - 4*a^2*b^2*c*d)*x)*sqr
t(a)*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) - 2*(3*a^3*b*c^2*x^3 + 2*(10*a^2*b^2*c^2 - 8*a^3*b*c*d + a
^4*d^2)*x^2 + 3*(5*a*b^3*c^2 - 4*a^2*b^2*c*d)*x)*sqrt((a*x + b)/x))/(a^6*b*x^2 + 2*a^5*b^2*x + a^4*b^3), 1/3*(
3*(5*b^4*c^2 - 4*a*b^3*c*d + (5*a^2*b^2*c^2 - 4*a^3*b*c*d)*x^2 + 2*(5*a*b^3*c^2 - 4*a^2*b^2*c*d)*x)*sqrt(-a)*a
rctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3*a^3*b*c^2*x^3 + 2*(10*a^2*b^2*c^2 - 8*a^3*b*c*d + a^4*d^2)*x^2 + 3*(5
*a*b^3*c^2 - 4*a^2*b^2*c*d)*x)*sqrt((a*x + b)/x))/(a^6*b*x^2 + 2*a^5*b^2*x + a^4*b^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x + d\right )^{2}}{x^{2} \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)**2/(a+b/x)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.20202, size = 217, normalized size = 1.78 \begin{align*} -\frac{1}{3} \, b{\left (\frac{3 \, c^{2} \sqrt{\frac{a x + b}{x}}}{{\left (a - \frac{a x + b}{x}\right )} a^{3}} - \frac{3 \,{\left (5 \, b c^{2} - 4 \, a c d\right )} \arctan \left (\frac{\sqrt{\frac{a x + b}{x}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3} b} - \frac{2 \,{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + \frac{6 \,{\left (a x + b\right )} b^{2} c^{2}}{x} - \frac{6 \,{\left (a x + b\right )} a b c d}{x}\right )} x}{{\left (a x + b\right )} a^{3} b^{2} \sqrt{\frac{a x + b}{x}}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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