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]

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

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Rubi [A]  time = 0.20, antiderivative size = 163, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.25, 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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fricas [B]  time = 0.64, size = 2493, normalized size = 15.29 \[ \text {result too large to display} \]

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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giac [B]  time = 0.76, size = 654, normalized size = 4.01 \[ \frac {\frac {{\left ({\left (b^{3} c - a^{2} b\right )}^{2} b^{4} c^{\frac {3}{2}} d^{2} - 2 \, {\left (a b^{6} c^{2} - a^{3} b^{4} c\right )} d^{2} {\left | -b^{3} c + a^{2} b \right |} + {\left (a^{2} b^{8} c^{\frac {5}{2}} - 2 \, a^{4} b^{6} c^{\frac {3}{2}} + a^{6} b^{4} \sqrt {c}\right )} d^{2}\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-\frac {a b^{2} c - a^{3} + \sqrt {{\left (a b^{2} c - a^{3}\right )}^{2} + {\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} {\left (b^{2} c - a^{2}\right )}}}{b^{2} c - a^{2}}}}\right )}{{\left (b^{5} c^{\frac {7}{2}} + a b^{4} c^{3} - 2 \, a^{2} b^{3} c^{\frac {5}{2}} - 2 \, a^{3} b^{2} c^{2} + a^{4} b c^{\frac {3}{2}} + a^{5} c\right )} \sqrt {b \sqrt {c} - a} {\left | -b^{3} c + a^{2} b \right |}} + \frac {{\left ({\left (b^{3} c - a^{2} b\right )}^{2} b^{4} c^{\frac {3}{2}} d^{2} + 2 \, {\left (a b^{6} c^{2} - a^{3} b^{4} c\right )} d^{2} {\left | -b^{3} c + a^{2} b \right |} + {\left (a^{2} b^{8} c^{\frac {5}{2}} - 2 \, a^{4} b^{6} c^{\frac {3}{2}} + a^{6} b^{4} \sqrt {c}\right )} d^{2}\right )} \arctan \left (\frac {\sqrt {\sqrt {d x + c} b + a}}{\sqrt {-\frac {a b^{2} c - a^{3} - \sqrt {{\left (a b^{2} c - a^{3}\right )}^{2} + {\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} {\left (b^{2} c - a^{2}\right )}}}{b^{2} c - a^{2}}}}\right )}{{\left (b^{5} c^{\frac {7}{2}} - a b^{4} c^{3} - 2 \, a^{2} b^{3} c^{\frac {5}{2}} + 2 \, a^{3} b^{2} c^{2} + a^{4} b c^{\frac {3}{2}} - a^{5} c\right )} \sqrt {-b \sqrt {c} - a} {\left | -b^{3} c + a^{2} b \right |}} + \frac {2 \, {\left ({\left (\sqrt {d x + c} b + a\right )}^{\frac {3}{2}} b^{4} d^{2} - 2 \, \sqrt {\sqrt {d x + c} b + a} a b^{4} d^{2}\right )}}{{\left (b^{2} c - {\left (\sqrt {d x + c} b + a\right )}^{2} + 2 \, {\left (\sqrt {d x + c} b + a\right )} a - a^{2}\right )} {\left (b^{2} c - a^{2}\right )}}}{2 \, b^{2} d} \]

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]

1/2*(((b^3*c - a^2*b)^2*b^4*c^(3/2)*d^2 - 2*(a*b^6*c^2 - a^3*b^4*c)*d^2*abs(-b^3*c + a^2*b) + (a^2*b^8*c^(5/2)
- 2*a^4*b^6*c^(3/2) + a^6*b^4*sqrt(c))*d^2)*arctan(sqrt(sqrt(d*x + c)*b + a)/sqrt(-(a*b^2*c - a^3 + sqrt((a*b
^2*c - a^3)^2 + (b^4*c^2 - 2*a^2*b^2*c + a^4)*(b^2*c - a^2)))/(b^2*c - a^2)))/((b^5*c^(7/2) + a*b^4*c^3 - 2*a^
2*b^3*c^(5/2) - 2*a^3*b^2*c^2 + a^4*b*c^(3/2) + a^5*c)*sqrt(b*sqrt(c) - a)*abs(-b^3*c + a^2*b)) + ((b^3*c - a^
2*b)^2*b^4*c^(3/2)*d^2 + 2*(a*b^6*c^2 - a^3*b^4*c)*d^2*abs(-b^3*c + a^2*b) + (a^2*b^8*c^(5/2) - 2*a^4*b^6*c^(3
/2) + a^6*b^4*sqrt(c))*d^2)*arctan(sqrt(sqrt(d*x + c)*b + a)/sqrt(-(a*b^2*c - a^3 - sqrt((a*b^2*c - a^3)^2 + (
b^4*c^2 - 2*a^2*b^2*c + a^4)*(b^2*c - a^2)))/(b^2*c - a^2)))/((b^5*c^(7/2) - a*b^4*c^3 - 2*a^2*b^3*c^(5/2) + 2
*a^3*b^2*c^2 + a^4*b*c^(3/2) - a^5*c)*sqrt(-b*sqrt(c) - a)*abs(-b^3*c + a^2*b)) + 2*((sqrt(d*x + c)*b + a)^(3/
2)*b^4*d^2 - 2*sqrt(sqrt(d*x + c)*b + a)*a*b^4*d^2)/((b^2*c - (sqrt(d*x + c)*b + a)^2 + 2*(sqrt(d*x + c)*b + a
)*a - a^2)*(b^2*c - a^2)))/(b^2*d)

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maple [B]  time = 0.03, size = 265, normalized size = 1.63 \[ \frac {2 \sqrt {b^{2} c}\, d \arctan \left (\frac {\sqrt {a +\sqrt {d x +c}\, b}}{\sqrt {-a -\sqrt {b^{2} c}}}\right )}{\left (-4 a -4 \sqrt {b^{2} c}\right ) \sqrt {-a -\sqrt {b^{2} c}}\, c}-\frac {2 \sqrt {b^{2} c}\, d \arctan \left (\frac {\sqrt {a +\sqrt {d x +c}\, b}}{\sqrt {-a +\sqrt {b^{2} c}}}\right )}{\left (-4 a +4 \sqrt {b^{2} c}\right ) \sqrt {-a +\sqrt {b^{2} c}}\, c}-\frac {2 \sqrt {b^{2} c}\, \sqrt {a +\sqrt {d x +c}\, b}\, d}{\left (-4 a -4 \sqrt {b^{2} c}\right ) \left (-\sqrt {d x +c}\, b +\sqrt {b^{2} c}\right ) c}-\frac {2 \sqrt {b^{2} c}\, \sqrt {a +\sqrt {d x +c}\, b}\, d}{\left (-4 a +4 \sqrt {b^{2} c}\right ) \left (\sqrt {d x +c}\, b +\sqrt {b^{2} c}\right ) c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\sqrt {d x + c} b + a} x^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,\sqrt {a+b\,\sqrt {c+d\,x}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \sqrt {a + b \sqrt {c + d x}}}\, dx \]

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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