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

Optimal. Leaf size=80 \[ \frac{a b d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{2 c^{3/2}}-\frac{\left (a+b \sqrt{c+d x}\right )^2}{2 x^2}-\frac{b d \left (a \sqrt{c+d x}+b c\right )}{2 c x} \]

[Out]

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

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Rubi [A]  time = 0.073545, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {371, 1398, 821, 12, 639, 207} \[ \frac{a b d^2 \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{2 c^{3/2}}-\frac{\left (a+b \sqrt{c+d x}\right )^2}{2 x^2}-\frac{b d \left (a \sqrt{c+d x}+b c\right )}{2 c x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[c + d*x])^2/x^3,x]

[Out]

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

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 0.356943, size = 221, normalized size = 2.76 \[ \frac{-\frac{2 \sqrt{c} \left (a^4 b^2 \left (-c^2+2 c d x+3 d^2 x^2\right )-a^2 b^4 c \left (c^2+4 c d x+2 d^2 x^2\right )-2 a^3 b^3 c \sqrt{c+d x} (2 c+d x)+a^5 b \sqrt{c+d x} (2 c+d x)+a^6 c+a b^5 c^2 \sqrt{c+d x} (2 c+d x)+b^6 c^2 (c+d x)^2\right )}{x^2 \left (a^2-b^2 c\right )^2}-a b d^2 \log \left (\sqrt{c}-\sqrt{c+d x}\right )+a b d^2 \log \left (\sqrt{c+d x}+\sqrt{c}\right )}{4 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 81, normalized size = 1. \begin{align*}{b}^{2} \left ( -{\frac{d}{x}}-{\frac{c}{2\,{x}^{2}}} \right ) +4\,ab{d}^{2} \left ({\frac{1}{{d}^{2}{x}^{2}} \left ( -1/8\,{\frac{ \left ( dx+c \right ) ^{3/2}}{c}}-1/8\,\sqrt{dx+c} \right ) }+1/8\,{\frac{1}{{c}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) } \right ) -{\frac{{a}^{2}}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(d*x+c)^(1/2))^2/x^3,x)

[Out]

b^2*(-d/x-1/2*c/x^2)+4*a*b*d^2*((-1/8/c*(d*x+c)^(3/2)-1/8*(d*x+c)^(1/2))/d^2/x^2+1/8/c^(3/2)*arctanh((d*x+c)^(
1/2)/c^(1/2)))-1/2*a^2/x^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((a+b*(d*x+c)^(1/2))^2/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72163, size = 427, normalized size = 5.34 \begin{align*} \left [\frac{a b \sqrt{c} d^{2} x^{2} \log \left (\frac{d x + 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) - 4 \, b^{2} c^{2} d x - 2 \, b^{2} c^{3} - 2 \, a^{2} c^{2} - 2 \,{\left (a b c d x + 2 \, a b c^{2}\right )} \sqrt{d x + c}}{4 \, c^{2} x^{2}}, -\frac{a b \sqrt{-c} d^{2} x^{2} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) + 2 \, b^{2} c^{2} d x + b^{2} c^{3} + a^{2} c^{2} +{\left (a b c d x + 2 \, a b c^{2}\right )} \sqrt{d x + c}}{2 \, c^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 119.286, size = 292, normalized size = 3.65 \begin{align*} - \frac{a^{2}}{2 x^{2}} - \frac{20 a b c^{2} d^{2} \sqrt{c + d x}}{- 8 c^{4} - 16 c^{3} d x + 8 c^{2} \left (c + d x\right )^{2}} + \frac{12 a b c d^{2} \left (c + d x\right )^{\frac{3}{2}}}{- 8 c^{4} - 16 c^{3} d x + 8 c^{2} \left (c + d x\right )^{2}} + \frac{3 a b c d^{2} \sqrt{\frac{1}{c^{5}}} \log{\left (- c^{3} \sqrt{\frac{1}{c^{5}}} + \sqrt{c + d x} \right )}}{4} - \frac{3 a b c d^{2} \sqrt{\frac{1}{c^{5}}} \log{\left (c^{3} \sqrt{\frac{1}{c^{5}}} + \sqrt{c + d x} \right )}}{4} - a b d^{2} \sqrt{\frac{1}{c^{3}}} \log{\left (- c^{2} \sqrt{\frac{1}{c^{3}}} + \sqrt{c + d x} \right )} + a b d^{2} \sqrt{\frac{1}{c^{3}}} \log{\left (c^{2} \sqrt{\frac{1}{c^{3}}} + \sqrt{c + d x} \right )} - \frac{2 a b d \sqrt{c + d x}}{c x} - \frac{b^{2} c}{2 x^{2}} - \frac{b^{2} d}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(d*x+c)**(1/2))**2/x**3,x)

[Out]

-a**2/(2*x**2) - 20*a*b*c**2*d**2*sqrt(c + d*x)/(-8*c**4 - 16*c**3*d*x + 8*c**2*(c + d*x)**2) + 12*a*b*c*d**2*
(c + d*x)**(3/2)/(-8*c**4 - 16*c**3*d*x + 8*c**2*(c + d*x)**2) + 3*a*b*c*d**2*sqrt(c**(-5))*log(-c**3*sqrt(c**
(-5)) + sqrt(c + d*x))/4 - 3*a*b*c*d**2*sqrt(c**(-5))*log(c**3*sqrt(c**(-5)) + sqrt(c + d*x))/4 - a*b*d**2*sqr
t(c**(-3))*log(-c**2*sqrt(c**(-3)) + sqrt(c + d*x)) + a*b*d**2*sqrt(c**(-3))*log(c**2*sqrt(c**(-3)) + sqrt(c +
d*x)) - 2*a*b*d*sqrt(c + d*x)/(c*x) - b**2*c/(2*x**2) - b**2*d/x

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Giac [A]  time = 1.16749, size = 170, normalized size = 2.12 \begin{align*} -\frac{\frac{a b d^{3} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{\sqrt{-c} c} + \frac{b^{2} c d^{3} - a^{2} d^{3}}{c^{2}} + \frac{2 \,{\left (d x + c\right )} b^{2} c d^{3} - b^{2} c^{2} d^{3} +{\left (d x + c\right )}^{\frac{3}{2}} a b d^{3} + \sqrt{d x + c} a b c d^{3} + a^{2} c d^{3}}{c d^{2} x^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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