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]

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

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Rubi [A]  time = 0.25, antiderivative size = 202, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 \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*}

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Mathematica [A]  time = 0.76, size = 230, normalized size = 1.14 \[ \frac {\frac {\sqrt {c} \left (2 b^2 \sqrt {c} d \left (3 a^2+b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )+\frac {\sqrt {c} \left (a^2-b^2 c\right ) \left (a^3-a^2 b \sqrt {c+d x}-a b^2 (c+4 d x)+b^3 c \sqrt {c+d x}\right )}{x \left (a+b \sqrt {c+d x}\right )}-b d \left (a+b \sqrt {c}\right )^3 \log \left (\sqrt {c+d x}+\sqrt {c}\right )\right )}{\left (a^2-b^2 c\right )^2}+\frac {\left (a b \sqrt {c} d-b^2 c d\right ) \log \left (\sqrt {c}-\sqrt {c+d x}\right )}{\left (a+b \sqrt {c}\right )^2}}{c \left (b^2 c-a^2\right )} \]

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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fricas [B]  time = 0.86, size = 854, normalized size = 4.23 \[ \left [-\frac {b^{6} c^{4} - a^{2} b^{4} c^{3} - a^{4} b^{2} c^{2} + a^{6} c + {\left (b^{6} c^{3} + 2 \, a^{2} b^{4} c^{2} - 3 \, a^{4} b^{2} c\right )} d x - {\left ({\left (3 \, a b^{5} c + a^{3} b^{3}\right )} d^{2} x^{2} + {\left (3 \, a b^{5} c^{2} - 2 \, a^{3} b^{3} c - a^{5} b\right )} d x\right )} \sqrt {c} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left ({\left (b^{6} c^{2} + 3 \, a^{2} b^{4} c\right )} d^{2} x^{2} + {\left (b^{6} c^{3} + 2 \, a^{2} b^{4} c^{2} - 3 \, a^{4} b^{2} c\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) + {\left ({\left (b^{6} c^{2} + 3 \, a^{2} b^{4} c\right )} d^{2} x^{2} + {\left (b^{6} c^{3} + 2 \, a^{2} b^{4} c^{2} - 3 \, a^{4} b^{2} c\right )} d x\right )} \log \relax (x) - 2 \, {\left (a b^{5} c^{3} - 2 \, a^{3} b^{3} c^{2} + a^{5} b c + 2 \, {\left (a b^{5} c^{2} - a^{3} b^{3} c\right )} d x\right )} \sqrt {d x + c}}{{\left (b^{8} c^{4} - 3 \, a^{2} b^{6} c^{3} + 3 \, a^{4} b^{4} c^{2} - a^{6} b^{2} c\right )} d x^{2} + {\left (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\right )} x}, -\frac {b^{6} c^{4} - a^{2} b^{4} c^{3} - a^{4} b^{2} c^{2} + a^{6} c + {\left (b^{6} c^{3} + 2 \, a^{2} b^{4} c^{2} - 3 \, a^{4} b^{2} c\right )} d x - 2 \, {\left ({\left (3 \, a b^{5} c + a^{3} b^{3}\right )} d^{2} x^{2} + {\left (3 \, a b^{5} c^{2} - 2 \, a^{3} b^{3} c - a^{5} b\right )} d x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c}}{c}\right ) - 2 \, {\left ({\left (b^{6} c^{2} + 3 \, a^{2} b^{4} c\right )} d^{2} x^{2} + {\left (b^{6} c^{3} + 2 \, a^{2} b^{4} c^{2} - 3 \, a^{4} b^{2} c\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) + {\left ({\left (b^{6} c^{2} + 3 \, a^{2} b^{4} c\right )} d^{2} x^{2} + {\left (b^{6} c^{3} + 2 \, a^{2} b^{4} c^{2} - 3 \, a^{4} b^{2} c\right )} d x\right )} \log \relax (x) - 2 \, {\left (a b^{5} c^{3} - 2 \, a^{3} b^{3} c^{2} + a^{5} b c + 2 \, {\left (a b^{5} c^{2} - a^{3} b^{3} c\right )} d x\right )} \sqrt {d x + c}}{{\left (b^{8} c^{4} - 3 \, a^{2} b^{6} c^{3} + 3 \, a^{4} b^{4} c^{2} - a^{6} b^{2} c\right )} d x^{2} + {\left (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\right )} x}\right ] \]

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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giac [A]  time = 0.41, size = 311, normalized size = 1.54 \[ -\frac {{\left (b^{4} c d + 3 \, a^{2} b^{2} d\right )} \log \left (-d x\right )}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}} + \frac {2 \, {\left (b^{5} c d + 3 \, a^{2} b^{3} d\right )} \log \left ({\left | -\sqrt {d x + c} b - a \right |}\right )}{b^{7} c^{3} - 3 \, a^{2} b^{5} c^{2} + 3 \, a^{4} b^{3} c - a^{6} b} + \frac {2 \, {\left (3 \, a b^{3} c d + a^{3} b d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{{\left (b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}\right )} \sqrt {-c}} - \frac {\sqrt {d x + c} b^{3} c d - 4 \, {\left (d x + c\right )} a b^{2} d + 3 \, a b^{2} c d - \sqrt {d x + c} a^{2} b d + a^{3} d}{{\left (b^{4} c^{2} - 2 \, a^{2} b^{2} c + a^{4}\right )} {\left ({\left (d x + c\right )}^{\frac {3}{2}} b - \sqrt {d x + c} b c + {\left (d x + c\right )} a - a c\right )}} \]

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]

-(b^4*c*d + 3*a^2*b^2*d)*log(-d*x)/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6) + 2*(b^5*c*d + 3*a^2*b^3*d)*l
og(abs(-sqrt(d*x + c)*b - a))/(b^7*c^3 - 3*a^2*b^5*c^2 + 3*a^4*b^3*c - a^6*b) + 2*(3*a*b^3*c*d + a^3*b*d)*arct
an(sqrt(d*x + c)/sqrt(-c))/((b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6)*sqrt(-c)) - (sqrt(d*x + c)*b^3*c*d -
4*(d*x + c)*a*b^2*d + 3*a*b^2*c*d - sqrt(d*x + c)*a^2*b*d + a^3*d)/((b^4*c^2 - 2*a^2*b^2*c + a^4)*((d*x + c)^
(3/2)*b - sqrt(d*x + c)*b*c + (d*x + c)*a - a*c))

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maple [A]  time = 0.02, size = 312, normalized size = 1.54 \[ \frac {b^{4} c d \ln \left (d x \right )}{\left (-b^{2} c +a^{2}\right )^{3}}-\frac {2 b^{4} c d \ln \left (a +\sqrt {d x +c}\, b \right )}{\left (-b^{2} c +a^{2}\right )^{3}}+\frac {6 a \,b^{3} \sqrt {c}\, d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{3}}+\frac {3 a^{2} b^{2} d \ln \left (d x \right )}{\left (-b^{2} c +a^{2}\right )^{3}}-\frac {6 a^{2} b^{2} d \ln \left (a +\sqrt {d x +c}\, b \right )}{\left (-b^{2} c +a^{2}\right )^{3}}+\frac {2 a^{3} b d \arctanh \left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{\left (-b^{2} c +a^{2}\right )^{3} \sqrt {c}}+\frac {b^{4} c^{2}}{\left (-b^{2} c +a^{2}\right )^{3} x}-\frac {2 \sqrt {d x +c}\, a \,b^{3} c}{\left (-b^{2} c +a^{2}\right )^{3} x}+\frac {2 a \,b^{2} d}{\left (-b^{2} c +a^{2}\right )^{2} \left (a +\sqrt {d x +c}\, b \right )}-\frac {a^{4}}{\left (-b^{2} c +a^{2}\right )^{3} x}+\frac {2 \sqrt {d x +c}\, a^{3} b}{\left (-b^{2} c +a^{2}\right )^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*b^4*c^2-1/
(-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)*b^2*a^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+2*a*b
^2*d/(-b^2*c+a^2)^2/(a+(d*x+c)^(1/2)*b)-2*d*b^4/(-b^2*c+a^2)^3*ln(a+(d*x+c)^(1/2)*b)*c-6*d*b^2/(-b^2*c+a^2)^3*
ln(a+(d*x+c)^(1/2)*b)*a^2

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maxima [A]  time = 2.01, size = 367, normalized size = 1.82 \[ -d {\left (\frac {{\left (b^{4} c + 3 \, a^{2} b^{2}\right )} \log \left (d x\right )}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}} - \frac {2 \, {\left (b^{4} c + 3 \, a^{2} b^{2}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}} - \frac {{\left (3 \, a b^{3} c + a^{3} b\right )} \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{{\left (b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6}\right )} \sqrt {c}} + \frac {4 \, {\left (d x + c\right )} a b^{2} - 3 \, a b^{2} c - a^{3} - {\left (b^{3} c - a^{2} b\right )} \sqrt {d x + c}}{a b^{4} c^{3} - 2 \, a^{3} b^{2} c^{2} + a^{5} c - {\left (b^{5} c^{2} - 2 \, a^{2} b^{3} c + a^{4} b\right )} {\left (d x + c\right )}^{\frac {3}{2}} - {\left (a b^{4} c^{2} - 2 \, a^{3} b^{2} c + a^{5}\right )} {\left (d x + c\right )} + {\left (b^{5} c^{3} - 2 \, a^{2} b^{3} c^{2} + a^{4} b c\right )} \sqrt {d x + c}}\right )} \]

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]

-d*((b^4*c + 3*a^2*b^2)*log(d*x)/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6) - 2*(b^4*c + 3*a^2*b^2)*log(sqr
t(d*x + c)*b + a)/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6) - (3*a*b^3*c + a^3*b)*log((sqrt(d*x + c) - sqr
t(c))/(sqrt(d*x + c) + sqrt(c)))/((b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6)*sqrt(c)) + (4*(d*x + c)*a*b^2
- 3*a*b^2*c - a^3 - (b^3*c - a^2*b)*sqrt(d*x + c))/(a*b^4*c^3 - 2*a^3*b^2*c^2 + a^5*c - (b^5*c^2 - 2*a^2*b^3*c
+ a^4*b)*(d*x + c)^(3/2) - (a*b^4*c^2 - 2*a^3*b^2*c + a^5)*(d*x + c) + (b^5*c^3 - 2*a^2*b^3*c^2 + a^4*b*c)*sq
rt(d*x + c)))

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mupad [B]  time = 0.73, size = 275, normalized size = 1.36 \[ \frac {b\,d\,\ln \left (\sqrt {c+d\,x}+\sqrt {c}\right )}{a^3\,\sqrt {c}-b^3\,c^2+3\,a\,b^2\,c^{3/2}-3\,a^2\,b\,c}-\frac {\frac {a\,d\,\left (a^2+3\,c\,b^2\right )}{{\left (b^2\,c-a^2\right )}^2}+\frac {b\,d\,\sqrt {c+d\,x}}{b^2\,c-a^2}-\frac {4\,a\,b^2\,d\,\left (c+d\,x\right )}{a^4-2\,a^2\,b^2\,c+b^4\,c^2}}{b\,{\left (c+d\,x\right )}^{3/2}-a\,c+a\,\left (c+d\,x\right )-b\,c\,\sqrt {c+d\,x}}-\frac {b\,d\,\ln \left (\sqrt {c+d\,x}-\sqrt {c}\right )}{a^3\,\sqrt {c}+b^3\,c^2+3\,a\,b^2\,c^{3/2}+3\,a^2\,b\,c}-\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (\frac {6\,b^2\,d}{{\left (b^2\,c-a^2\right )}^2}-\frac {8\,b^4\,c\,d}{{\left (b^2\,c-a^2\right )}^3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*d*log((c + d*x)^(1/2) + c^(1/2)))/(a^3*c^(1/2) - b^3*c^2 + 3*a*b^2*c^(3/2) - 3*a^2*b*c) - ((a*d*(3*b^2*c +
a^2))/(b^2*c - a^2)^2 + (b*d*(c + d*x)^(1/2))/(b^2*c - a^2) - (4*a*b^2*d*(c + d*x))/(a^4 + b^4*c^2 - 2*a^2*b^2
*c))/(b*(c + d*x)^(3/2) - a*c + a*(c + d*x) - b*c*(c + d*x)^(1/2)) - (b*d*log((c + d*x)^(1/2) - c^(1/2)))/(a^3
*c^(1/2) + b^3*c^2 + 3*a*b^2*c^(3/2) + 3*a^2*b*c) - log(a + b*(c + d*x)^(1/2))*((6*b^2*d)/(b^2*c - a^2)^2 - (8
*b^4*c*d)/(b^2*c - a^2)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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