3.476 \(\int \frac{1}{x^2 (a+b x)^2 \sqrt{c+d x}} \, dx\)

Optimal. Leaf size=164 \[ -\frac{b^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 (b c-a d)^{3/2}}+\frac{(a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3 c^{3/2}}-\frac{b \sqrt{c+d x} (2 b c-a d)}{a^2 c (a+b x) (b c-a d)}-\frac{\sqrt{c+d x}}{a c x (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.201657, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {103, 151, 156, 63, 208} \[ -\frac{b^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 (b c-a d)^{3/2}}+\frac{(a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3 c^{3/2}}-\frac{b \sqrt{c+d x} (2 b c-a d)}{a^2 c (a+b x) (b c-a d)}-\frac{\sqrt{c+d x}}{a c x (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x)^2*Sqrt[c + d*x]),x]

[Out]

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

Rule 103

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

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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{1}{x^2 (a+b x)^2 \sqrt{c+d x}} \, dx &=-\frac{\sqrt{c+d x}}{a c x (a+b x)}-\frac{\int \frac{\frac{1}{2} (4 b c+a d)+\frac{3 b d x}{2}}{x (a+b x)^2 \sqrt{c+d x}} \, dx}{a c}\\ &=-\frac{b (2 b c-a d) \sqrt{c+d x}}{a^2 c (b c-a d) (a+b x)}-\frac{\sqrt{c+d x}}{a c x (a+b x)}-\frac{\int \frac{\frac{1}{2} (b c-a d) (4 b c+a d)+\frac{1}{2} b d (2 b c-a d) x}{x (a+b x) \sqrt{c+d x}} \, dx}{a^2 c (b c-a d)}\\ &=-\frac{b (2 b c-a d) \sqrt{c+d x}}{a^2 c (b c-a d) (a+b x)}-\frac{\sqrt{c+d x}}{a c x (a+b x)}+\frac{\left (b^2 (4 b c-5 a d)\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 a^3 (b c-a d)}-\frac{(4 b c+a d) \int \frac{1}{x \sqrt{c+d x}} \, dx}{2 a^3 c}\\ &=-\frac{b (2 b c-a d) \sqrt{c+d x}}{a^2 c (b c-a d) (a+b x)}-\frac{\sqrt{c+d x}}{a c x (a+b x)}+\frac{\left (b^2 (4 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^3 d (b c-a d)}-\frac{(4 b c+a d) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^3 c d}\\ &=-\frac{b (2 b c-a d) \sqrt{c+d x}}{a^2 c (b c-a d) (a+b x)}-\frac{\sqrt{c+d x}}{a c x (a+b x)}+\frac{(4 b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3 c^{3/2}}-\frac{b^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.490817, size = 148, normalized size = 0.9 \[ \frac{\frac{a \sqrt{c+d x} \left (a^2 d+a b (d x-c)-2 b^2 c x\right )}{x (a+b x) (b c-a d)}+\frac{b^{3/2} c (5 a d-4 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{3/2}}+\frac{(a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\sqrt{c}}}{a^3 c} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*x)^2*Sqrt[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.016, size = 202, normalized size = 1.2 \begin{align*} -{\frac{1}{{a}^{2}cx}\sqrt{dx+c}}+{\frac{d}{{a}^{2}}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ){c}^{-{\frac{3}{2}}}}+4\,{\frac{b}{{a}^{3}\sqrt{c}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+{\frac{{b}^{2}d}{{a}^{2} \left ( ad-bc \right ) \left ( bdx+ad \right ) }\sqrt{dx+c}}+5\,{\frac{{b}^{2}d}{{a}^{2} \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-4\,{\frac{{b}^{3}c}{{a}^{3} \left ( ad-bc \right ) \sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 10.1599, size = 2396, normalized size = 14.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [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(1/x**2/(b*x+a)**2/(d*x+c)**(1/2),x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.17904, size = 319, normalized size = 1.95 \begin{align*} \frac{{\left (4 \, b^{3} c - 5 \, a b^{2} d\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (a^{3} b c - a^{4} d\right )} \sqrt{-b^{2} c + a b d}} - \frac{2 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{2} c d - 2 \, \sqrt{d x + c} b^{2} c^{2} d -{\left (d x + c\right )}^{\frac{3}{2}} a b d^{2} + 2 \, \sqrt{d x + c} a b c d^{2} - \sqrt{d x + c} a^{2} d^{3}}{{\left (a^{2} b c^{2} - a^{3} c d\right )}{\left ({\left (d x + c\right )}^{2} b - 2 \,{\left (d x + c\right )} b c + b c^{2} +{\left (d x + c\right )} a d - a c d\right )}} - \frac{{\left (4 \, b c + a d\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{a^{3} \sqrt{-c} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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