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

Optimal. Leaf size=163 \[ -\frac{\sqrt{a+b \sqrt{c+d x}} \left (a-b \sqrt{c+d x}\right )}{x \left (a^2-b^2 c\right )}-\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a-b \sqrt{c}}}\right )}{2 \sqrt{c} \left (a-b \sqrt{c}\right )^{3/2}}+\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a+b \sqrt{c}}}\right )}{2 \sqrt{c} \left (a+b \sqrt{c}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.203434, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {371, 1398, 823, 827, 1166, 207} \[ -\frac{\sqrt{a+b \sqrt{c+d x}} \left (a-b \sqrt{c+d x}\right )}{x \left (a^2-b^2 c\right )}-\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a-b \sqrt{c}}}\right )}{2 \sqrt{c} \left (a-b \sqrt{c}\right )^{3/2}}+\frac{b d \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a+b \sqrt{c}}}\right )}{2 \sqrt{c} \left (a+b \sqrt{c}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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

Mathematica [A]  time = 0.257769, size = 216, normalized size = 1.33 \[ \frac{\sqrt{a-b \sqrt{c}} \left (b d x \left (a-b \sqrt{c}\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a+b \sqrt{c}}}\right )-2 \sqrt{c} \sqrt{a+b \sqrt{c}} \left (a-b \sqrt{c+d x}\right ) \sqrt{a+b \sqrt{c+d x}}\right )-b d x \left (a+b \sqrt{c}\right )^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c+d x}}}{\sqrt{a-b \sqrt{c}}}\right )}{2 \sqrt{c} x \sqrt{a-b \sqrt{c}} \sqrt{a+b \sqrt{c}} \left (a^2-b^2 c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.022, size = 265, normalized size = 1.6 \begin{align*} -2\,{\frac{d\sqrt{{b}^{2}c}\sqrt{a+b\sqrt{dx+c}}}{c \left ( 4\,\sqrt{{b}^{2}c}-4\,a \right ) \left ( b\sqrt{dx+c}+\sqrt{{b}^{2}c} \right ) }}-2\,{\frac{d\sqrt{{b}^{2}c}}{c \left ( 4\,\sqrt{{b}^{2}c}-4\,a \right ) \sqrt{\sqrt{{b}^{2}c}-a}}\arctan \left ({\frac{\sqrt{a+b\sqrt{dx+c}}}{\sqrt{\sqrt{{b}^{2}c}-a}}} \right ) }-2\,{\frac{d\sqrt{{b}^{2}c}\sqrt{a+b\sqrt{dx+c}}}{c \left ( -4\,\sqrt{{b}^{2}c}-4\,a \right ) \left ( -b\sqrt{dx+c}+\sqrt{{b}^{2}c} \right ) }}+2\,{\frac{d\sqrt{{b}^{2}c}}{c \left ( -4\,\sqrt{{b}^{2}c}-4\,a \right ) \sqrt{-\sqrt{{b}^{2}c}-a}}\arctan \left ({\frac{\sqrt{a+b\sqrt{dx+c}}}{\sqrt{-\sqrt{{b}^{2}c}-a}}} \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))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.9717, size = 4913, normalized size = 30.14 \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))^(1/2),x, algorithm="fricas")

[Out]

1/4*((b^2*c - a^2)*x*sqrt(-((3*a*b^4*c + a^3*b^2)*d^2 + (b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*sqrt
((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^
8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c))*log((b^6*c + 3*a^2*b
^4)*sqrt(sqrt(d*x + c)*b + a)*d^3 + (2*(a*b^6*c^2 + 3*a^3*b^4*c)*d^2 - (b^8*c^5 - 2*a^2*b^6*c^4 + 2*a^6*b^2*c^
2 - a^8*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4*b^8*c^5 - 20*a^6*
b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))*sqrt(-((3*a*b^4*c + a^3*b^2)*d^2 + (b^6*c^4 - 3*a^2*b^4*
c^3 + 3*a^4*b^2*c^2 - a^6*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4
*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c
^2 - a^6*c))) - (b^2*c - a^2)*x*sqrt(-((3*a*b^4*c + a^3*b^2)*d^2 + (b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 -
a^6*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4*b^8*c^5 - 20*a^6*b^6*
c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c))*log((b^6*
c + 3*a^2*b^4)*sqrt(sqrt(d*x + c)*b + a)*d^3 - (2*(a*b^6*c^2 + 3*a^3*b^4*c)*d^2 - (b^8*c^5 - 2*a^2*b^6*c^4 + 2
*a^6*b^2*c^2 - a^8*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4*b^8*c^
5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))*sqrt(-((3*a*b^4*c + a^3*b^2)*d^2 + (b^6*c^4 -
3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c
^6 + 15*a^4*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*c^4 - 3*a^2*b^4*c^3 +
3*a^4*b^2*c^2 - a^6*c))) + (b^2*c - a^2)*x*sqrt(-((3*a*b^4*c + a^3*b^2)*d^2 - (b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4
*b^2*c^2 - a^6*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4*b^8*c^5 -
20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)
)*log((b^6*c + 3*a^2*b^4)*sqrt(sqrt(d*x + c)*b + a)*d^3 + (2*(a*b^6*c^2 + 3*a^3*b^4*c)*d^2 + (b^8*c^5 - 2*a^2*
b^6*c^4 + 2*a^6*b^2*c^2 - a^8*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15
*a^4*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))*sqrt(-((3*a*b^4*c + a^3*b^2)*d^2 -
(b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6
*a^2*b^10*c^6 + 15*a^4*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*c^4 - 3*a^2
*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c))) - (b^2*c - a^2)*x*sqrt(-((3*a*b^4*c + a^3*b^2)*d^2 - (b^6*c^4 - 3*a^2*b^4*
c^3 + 3*a^4*b^2*c^2 - a^6*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4
*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c
^2 - a^6*c))*log((b^6*c + 3*a^2*b^4)*sqrt(sqrt(d*x + c)*b + a)*d^3 - (2*(a*b^6*c^2 + 3*a^3*b^4*c)*d^2 + (b^8*c
^5 - 2*a^2*b^6*c^4 + 2*a^6*b^2*c^2 - a^8*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b^12*c^7 - 6*a^2*b^
10*c^6 + 15*a^4*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))*sqrt(-((3*a*b^4*c + a^3
*b^2)*d^2 - (b^6*c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c)*sqrt((b^10*c^2 + 6*a^2*b^8*c + 9*a^4*b^6)*d^4/(b
^12*c^7 - 6*a^2*b^10*c^6 + 15*a^4*b^8*c^5 - 20*a^6*b^6*c^4 + 15*a^8*b^4*c^3 - 6*a^10*b^2*c^2 + a^12*c)))/(b^6*
c^4 - 3*a^2*b^4*c^3 + 3*a^4*b^2*c^2 - a^6*c))) - 4*sqrt(sqrt(d*x + c)*b + a)*(sqrt(d*x + c)*b - a))/((b^2*c -
a^2)*x)

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out