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

Optimal. Leaf size=202 \[ \frac{4 a b^2 d}{\left (a^2-b^2 c\right )^2 \left (a+b \sqrt{c+d x}\right )}-\frac{a-b \sqrt{c+d x}}{x \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}+\frac{b^2 d \log (x) \left (3 a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^3}-\frac{2 b^2 d \left (3 a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^3}+\frac{2 a b d \left (a^2+3 b^2 c\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\sqrt{c} \left (a^2-b^2 c\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.245105, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {371, 1398, 823, 801, 635, 206, 260} \[ \frac{4 a b^2 d}{\left (a^2-b^2 c\right )^2 \left (a+b \sqrt{c+d x}\right )}-\frac{a-b \sqrt{c+d x}}{x \left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}+\frac{b^2 d \log (x) \left (3 a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^3}-\frac{2 b^2 d \left (3 a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^3}+\frac{2 a b d \left (a^2+3 b^2 c\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\sqrt{c} \left (a^2-b^2 c\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

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

Rule 206

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.799447, size = 230, normalized size = 1.14 \[ \frac{\frac{\sqrt{c} \left (\frac{\sqrt{c} \left (a^2-b^2 c\right ) \left (-a^2 b \sqrt{c+d x}+a^3-a b^2 (c+4 d x)+b^3 c \sqrt{c+d x}\right )}{x \left (a+b \sqrt{c+d x}\right )}+2 b^2 \sqrt{c} d \left (3 a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )-b d \left (a+b \sqrt{c}\right )^3 \log \left (\sqrt{c+d x}+\sqrt{c}\right )\right )}{\left (a^2-b^2 c\right )^2}+\frac{\left (a b \sqrt{c} d-b^2 c d\right ) \log \left (\sqrt{c}-\sqrt{c+d x}\right )}{\left (a+b \sqrt{c}\right )^2}}{c \left (b^2 c-a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 312, normalized size = 1.5 \begin{align*} 2\,{\frac{a{b}^{2}d}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{2} \left ( a+b\sqrt{dx+c} \right ) }}-2\,{\frac{{b}^{4}d\ln \left ( a+b\sqrt{dx+c} \right ) c}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}}-6\,{\frac{{b}^{2}d\ln \left ( a+b\sqrt{dx+c} \right ){a}^{2}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}}-2\,{\frac{a\sqrt{dx+c}{b}^{3}c}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}x}}+2\,{\frac{\sqrt{dx+c}{a}^{3}b}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}x}}+{\frac{{c}^{2}{b}^{4}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}x}}-{\frac{{a}^{4}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}x}}+{\frac{d\ln \left ( dx \right ){b}^{4}c}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}}+3\,{\frac{d\ln \left ( dx \right ){a}^{2}{b}^{2}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}}+6\,{\frac{d\sqrt{c}a{b}^{3}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+2\,{\frac{bd{a}^{3}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{3}\sqrt{c}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*b^2*d/(-b^2*c+a^2)^2/(a+b*(d*x+c)^(1/2))-2*d*b^4/(-b^2*c+a^2)^3*ln(a+b*(d*x+c)^(1/2))*c-6*d*b^2/(-b^2*c+a^
2)^3*ln(a+b*(d*x+c)^(1/2))*a^2-2/(-b^2*c+a^2)^3/x*(d*x+c)^(1/2)*a*b^3*c+2/(-b^2*c+a^2)^3/x*(d*x+c)^(1/2)*a^3*b
+1/(-b^2*c+a^2)^3/x*c^2*b^4-1/(-b^2*c+a^2)^3/x*a^4+d/(-b^2*c+a^2)^3*ln(d*x)*b^4*c+3*d/(-b^2*c+a^2)^3*ln(d*x)*a
^2*b^2+6*d/(-b^2*c+a^2)^3*c^(1/2)*arctanh((d*x+c)^(1/2)/c^(1/2))*a*b^3+2*d/(-b^2*c+a^2)^3*b/c^(1/2)*arctanh((d
*x+c)^(1/2)/c^(1/2))*a^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(1/x^2/(a+b*(d*x+c)^(1/2))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.91541, size = 1724, normalized size = 8.53 \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/(a+b*(d*x+c)^(1/2))^2,x, algorithm="fricas")

[Out]

[-(b^6*c^4 - a^2*b^4*c^3 - a^4*b^2*c^2 + a^6*c + (b^6*c^3 + 2*a^2*b^4*c^2 - 3*a^4*b^2*c)*d*x - ((3*a*b^5*c + a
^3*b^3)*d^2*x^2 + (3*a*b^5*c^2 - 2*a^3*b^3*c - a^5*b)*d*x)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x
) - 2*((b^6*c^2 + 3*a^2*b^4*c)*d^2*x^2 + (b^6*c^3 + 2*a^2*b^4*c^2 - 3*a^4*b^2*c)*d*x)*log(sqrt(d*x + c)*b + a)
+ ((b^6*c^2 + 3*a^2*b^4*c)*d^2*x^2 + (b^6*c^3 + 2*a^2*b^4*c^2 - 3*a^4*b^2*c)*d*x)*log(x) - 2*(a*b^5*c^3 - 2*a
^3*b^3*c^2 + a^5*b*c + 2*(a*b^5*c^2 - a^3*b^3*c)*d*x)*sqrt(d*x + c))/((b^8*c^4 - 3*a^2*b^6*c^3 + 3*a^4*b^4*c^2
- a^6*b^2*c)*d*x^2 + (b^8*c^5 - 4*a^2*b^6*c^4 + 6*a^4*b^4*c^3 - 4*a^6*b^2*c^2 + a^8*c)*x), -(b^6*c^4 - a^2*b^
4*c^3 - a^4*b^2*c^2 + a^6*c + (b^6*c^3 + 2*a^2*b^4*c^2 - 3*a^4*b^2*c)*d*x - 2*((3*a*b^5*c + a^3*b^3)*d^2*x^2 +
(3*a*b^5*c^2 - 2*a^3*b^3*c - a^5*b)*d*x)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) - 2*((b^6*c^2 + 3*a^2*b^4*
c)*d^2*x^2 + (b^6*c^3 + 2*a^2*b^4*c^2 - 3*a^4*b^2*c)*d*x)*log(sqrt(d*x + c)*b + a) + ((b^6*c^2 + 3*a^2*b^4*c)*
d^2*x^2 + (b^6*c^3 + 2*a^2*b^4*c^2 - 3*a^4*b^2*c)*d*x)*log(x) - 2*(a*b^5*c^3 - 2*a^3*b^3*c^2 + a^5*b*c + 2*(a*
b^5*c^2 - a^3*b^3*c)*d*x)*sqrt(d*x + c))/((b^8*c^4 - 3*a^2*b^6*c^3 + 3*a^4*b^4*c^2 - a^6*b^2*c)*d*x^2 + (b^8*c
^5 - 4*a^2*b^6*c^4 + 6*a^4*b^4*c^3 - 4*a^6*b^2*c^2 + a^8*c)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (a + b \sqrt{c + d x}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*(a + b*sqrt(c + d*x))**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef