3.7.6 \(\int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x)^2 \sqrt {3+4 x^4}} \, dx\)

Optimal. Leaf size=268 \[ \frac {\left (\frac {1}{2}-\frac {i}{2}\right ) d \sqrt {\sqrt {3}-2 i x^2}}{\left (-\sqrt {3} d^2+2 i c^2\right ) (c+d x)}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) d \sqrt {\sqrt {3}+2 i x^2}}{\left (\sqrt {3} d^2+2 i c^2\right ) (c+d x)}+\frac {(1+i) c \tan ^{-1}\left (\frac {\sqrt {3} d+2 i c x}{\sqrt {\sqrt {3}-2 i x^2} \sqrt {-\sqrt {3} d^2+2 i c^2}}\right )}{\left (-\sqrt {3} d^2+2 i c^2\right )^{3/2}}+\frac {(1-i) c \tanh ^{-1}\left (\frac {\sqrt {3} d-2 i c x}{\sqrt {\sqrt {3}+2 i x^2} \sqrt {\sqrt {3} d^2+2 i c^2}}\right )}{\left (\sqrt {3} d^2+2 i c^2\right )^{3/2}} \]

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Rubi [A]  time = 0.31, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2133, 731, 725, 204, 206} \begin {gather*} \frac {\left (\frac {1}{2}-\frac {i}{2}\right ) d \sqrt {\sqrt {3}-2 i x^2}}{\left (-\sqrt {3} d^2+2 i c^2\right ) (c+d x)}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) d \sqrt {\sqrt {3}+2 i x^2}}{\left (\sqrt {3} d^2+2 i c^2\right ) (c+d x)}+\frac {(1+i) c \tan ^{-1}\left (\frac {\sqrt {3} d+2 i c x}{\sqrt {\sqrt {3}-2 i x^2} \sqrt {-\sqrt {3} d^2+2 i c^2}}\right )}{\left (-\sqrt {3} d^2+2 i c^2\right )^{3/2}}+\frac {(1-i) c \tanh ^{-1}\left (\frac {\sqrt {3} d-2 i c x}{\sqrt {\sqrt {3}+2 i x^2} \sqrt {\sqrt {3} d^2+2 i c^2}}\right )}{\left (\sqrt {3} d^2+2 i c^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)^2*Sqrt[3 + 4*x^4]),x]

[Out]

((1/2 - I/2)*d*Sqrt[Sqrt[3] - (2*I)*x^2])/(((2*I)*c^2 - Sqrt[3]*d^2)*(c + d*x)) - ((1/2 + I/2)*d*Sqrt[Sqrt[3]
+ (2*I)*x^2])/(((2*I)*c^2 + Sqrt[3]*d^2)*(c + d*x)) + ((1 + I)*c*ArcTan[(Sqrt[3]*d + (2*I)*c*x)/(Sqrt[(2*I)*c^
2 - Sqrt[3]*d^2]*Sqrt[Sqrt[3] - (2*I)*x^2])])/((2*I)*c^2 - Sqrt[3]*d^2)^(3/2) + ((1 - I)*c*ArcTanh[(Sqrt[3]*d
- (2*I)*c*x)/(Sqrt[(2*I)*c^2 + Sqrt[3]*d^2]*Sqrt[Sqrt[3] + (2*I)*x^2])])/((2*I)*c^2 + Sqrt[3]*d^2)^(3/2)

Rule 204

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

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 725

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

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 2133

Int[(((c_.) + (d_.)*(x_))^(m_.)*Sqrt[(b_.)*(x_)^2 + Sqrt[(a_) + (e_.)*(x_)^4]])/Sqrt[(a_) + (e_.)*(x_)^4], x_S
ymbol] :> Dist[(1 - I)/2, Int[(c + d*x)^m/Sqrt[Sqrt[a] - I*b*x^2], x], x] + Dist[(1 + I)/2, Int[(c + d*x)^m/Sq
rt[Sqrt[a] + I*b*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[e, b^2] && GtQ[a, 0]

Rubi steps

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

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Mathematica [F]  time = 0.10, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x)^2 \sqrt {3+4 x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)^2*Sqrt[3 + 4*x^4]),x]

[Out]

Integrate[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)^2*Sqrt[3 + 4*x^4]), x]

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IntegrateAlgebraic [C]  time = 5.17, size = 1609, normalized size = 6.00

result too large to display

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]/((c + d*x)^2*Sqrt[3 + 4*x^4]),x]

[Out]

(-(d*(-6*c^3 - 6*c*d^2*x^2 - 16*c^3*x^4)*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]) - d*x*(6*c^2*d + 6*d^3*x^2 + 16*c^2*d*
x^4)*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] + Sqrt[3 + 4*x^4]*(-(d*(-3*c*d^2 - 8*c^3*x^2)*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]
) - d*x*(3*d^3 + 8*c^2*d*x^2)*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]]))/(4*(4*c^4 + 3*d^4)*x^2*(-c^2 + d^2*x^2)*Sqrt[3 +
4*x^4] + (4*c^4 + 3*d^4)*(-c^2 + d^2*x^2)*(3 + 8*x^4)) + (8*c^5*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt
[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]])/((4*c^4 + 3*d^4)^(3/2)*Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]) - (6*c*d^4*ArcTan[
(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]])/((4*c^4 + 3*d^4)^(3/2)*Sqrt[-2*c^2 - Sq
rt[4*c^4 + 3*d^4]]) + (4*c^3*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]])/((4
*c^4 + 3*d^4)*Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]) - (8*c^5*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[-2*c^
2 + Sqrt[4*c^4 + 3*d^4]]])/((4*c^4 + 3*d^4)^(3/2)*Sqrt[-2*c^2 + Sqrt[4*c^4 + 3*d^4]]) + (6*c*d^4*ArcTan[(d*Sqr
t[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[-2*c^2 + Sqrt[4*c^4 + 3*d^4]]])/((4*c^4 + 3*d^4)^(3/2)*Sqrt[-2*c^2 + Sqrt[4*c
^4 + 3*d^4]]) + (4*c^3*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[-2*c^2 + Sqrt[4*c^4 + 3*d^4]]])/((4*c^4 +
3*d^4)*Sqrt[-2*c^2 + Sqrt[4*c^4 + 3*d^4]]) - RootSum[9*d^2 - 24*c^2*#1 - 6*d^2*#1^2 - 8*c^2*#1^3 + d^2*#1^4 &
, (128*c^4*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1] + 3*d^4*Log[2*x^2 + Sqrt[3 +
4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1] + 16*c^2*d^2*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 +
Sqrt[3 + 4*x^4]] - #1]*#1 + d^4*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1]*#1^2)/(6
*c^2 + 3*d^2*#1 + 6*c^2*#1^2 - d^2*#1^3) & ]/d^4 + RootSum[9*d^2 - 24*c^2*#1 - 6*d^2*#1^2 - 8*c^2*#1^3 + d^2*#
1^4 & , (512*c^8*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1] + 408*c^4*d^4*Log[2*x^2
+ Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1] + 9*d^8*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*
x^2 + Sqrt[3 + 4*x^4]] - #1] + 64*c^6*d^2*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1
]*#1 + 36*c^2*d^6*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1]*#1 + 8*c^4*d^4*Log[2*x
^2 + Sqrt[3 + 4*x^4] + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1]*#1^2 + 3*d^8*Log[2*x^2 + Sqrt[3 + 4*x^4] + 2*x*
Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - #1]*#1^2)/(6*c^2 + 3*d^2*#1 + 6*c^2*#1^2 - d^2*#1^3) & ]/(d^4*(4*c^4 + 3*d^4))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)^2/(4*x^4+3)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 \, x^{2} + \sqrt {4 \, x^{4} + 3}}}{\sqrt {4 \, x^{4} + 3} {\left (d x + c\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)^2/(4*x^4+3)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(2*x^2 + sqrt(4*x^4 + 3))/(sqrt(4*x^4 + 3)*(d*x + c)^2), x)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 x^{2}+\sqrt {4 x^{4}+3}}}{\left (d x +c \right )^{2} \sqrt {4 x^{4}+3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)^2/(4*x^4+3)^(1/2),x)

[Out]

int((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)^2/(4*x^4+3)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 \, x^{2} + \sqrt {4 \, x^{4} + 3}}}{\sqrt {4 \, x^{4} + 3} {\left (d x + c\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+(4*x^4+3)^(1/2))^(1/2)/(d*x+c)^2/(4*x^4+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(2*x^2 + sqrt(4*x^4 + 3))/(sqrt(4*x^4 + 3)*(d*x + c)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {2\,x^2+\sqrt {4\,x^4+3}}}{\sqrt {4\,x^4+3}\,{\left (c+d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2 + (4*x^4 + 3)^(1/2))^(1/2)/((4*x^4 + 3)^(1/2)*(c + d*x)^2),x)

[Out]

int((2*x^2 + (4*x^4 + 3)^(1/2))^(1/2)/((4*x^4 + 3)^(1/2)*(c + d*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {2 x^{2} + \sqrt {4 x^{4} + 3}}}{\left (c + d x\right )^{2} \sqrt {4 x^{4} + 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+(4*x**4+3)**(1/2))**(1/2)/(d*x+c)**2/(4*x**4+3)**(1/2),x)

[Out]

Integral(sqrt(2*x**2 + sqrt(4*x**4 + 3))/((c + d*x)**2*sqrt(4*x**4 + 3)), x)

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