3.10.13 \(\int \frac {\sqrt {2 x^2+\sqrt {3+4 x^4}}}{(c+d x) \sqrt {3+4 x^4}} \, dx\) [913]

Optimal. Leaf size=169 \[ \frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \tan ^{-1}\left (\frac {\sqrt {3} d+2 i c x}{\sqrt {2 i c^2-\sqrt {3} d^2} \sqrt {\sqrt {3}-2 i x^2}}\right )}{\sqrt {2 i c^2-\sqrt {3} d^2}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \tanh ^{-1}\left (\frac {\sqrt {3} d-2 i c x}{\sqrt {2 i c^2+\sqrt {3} d^2} \sqrt {\sqrt {3}+2 i x^2}}\right )}{\sqrt {2 i c^2+\sqrt {3} d^2}} \]

[Out]

(1/2-1/2*I)*arctan((2*I*c*x+d*3^(1/2))/(-2*I*x^2+3^(1/2))^(1/2)/(2*I*c^2-d^2*3^(1/2))^(1/2))/(2*I*c^2-d^2*3^(1
/2))^(1/2)-(1/2+1/2*I)*arctanh((-2*I*c*x+d*3^(1/2))/(2*I*x^2+3^(1/2))^(1/2)/(2*I*c^2+d^2*3^(1/2))^(1/2))/(2*I*
c^2+d^2*3^(1/2))^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2158, 739, 210, 212} \begin {gather*} \frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \text {ArcTan}\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 )}{\sqrt {-\sqrt {3} d^2+2 i c^2}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \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 )}{\sqrt {\sqrt {3} d^2+2 i c^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 210

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

Rule 212

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

Rule 739

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 2158

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 4.00, size = 531, normalized size = 3.14 \begin {gather*} \frac {-\sqrt {-2 c^2-\sqrt {4 c^4+3 d^4}} \tan ^{-1}\left (\frac {d \sqrt {2 x^2+\sqrt {3+4 x^4}}}{\sqrt {-2 c^2-\sqrt {4 c^4+3 d^4}}}\right )+\sqrt {-2 c^2+\sqrt {4 c^4+3 d^4}} \tan ^{-1}\left (\frac {d \sqrt {2 x^2+\sqrt {3+4 x^4}}}{\sqrt {-2 c^2+\sqrt {4 c^4+3 d^4}}}\right )+2 i c \sqrt {12 c^4+9 d^4} \text {RootSum}\left [12 d^2+16 i \sqrt {3} c^2 \text {$\#$1}+24 d^2 \text {$\#$1}+24 i \sqrt {3} c^2 \text {$\#$1}^2+24 d^2 \text {$\#$1}^2+8 i \sqrt {3} c^2 \text {$\#$1}^3+12 d^2 \text {$\#$1}^3+3 d^2 \text {$\#$1}^4\&,\frac {\log \left (2 x^2+\sqrt {3+4 x^4}\right )-\log \left (i \sqrt {3}-2 x^2+2 x \sqrt {2 x^2+\sqrt {3+4 x^4}}-2 x^2 \text {$\#$1}-\sqrt {3+4 x^4} (1+\text {$\#$1})\right )+\log \left (2 x^2+\sqrt {3+4 x^4}\right ) \text {$\#$1}-\log \left (i \sqrt {3}-2 x^2+2 x \sqrt {2 x^2+\sqrt {3+4 x^4}}-2 x^2 \text {$\#$1}-\sqrt {3+4 x^4} (1+\text {$\#$1})\right ) \text {$\#$1}}{4 i \sqrt {3} c^2+6 d^2+12 i \sqrt {3} c^2 \text {$\#$1}+12 d^2 \text {$\#$1}+6 i \sqrt {3} c^2 \text {$\#$1}^2+9 d^2 \text {$\#$1}^2+3 d^2 \text {$\#$1}^3}\&\right ]}{\sqrt {4 c^4+3 d^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^4]]*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[-2*c^2 - Sqrt[4*c^4 + 3*d^
4]]]) + Sqrt[-2*c^2 + Sqrt[4*c^4 + 3*d^4]]*ArcTan[(d*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]])/Sqrt[-2*c^2 + Sqrt[4*c^4 +
3*d^4]]] + (2*I)*c*Sqrt[12*c^4 + 9*d^4]*RootSum[12*d^2 + (16*I)*Sqrt[3]*c^2*#1 + 24*d^2*#1 + (24*I)*Sqrt[3]*c
^2*#1^2 + 24*d^2*#1^2 + (8*I)*Sqrt[3]*c^2*#1^3 + 12*d^2*#1^3 + 3*d^2*#1^4 & , (Log[2*x^2 + Sqrt[3 + 4*x^4]] -
Log[I*Sqrt[3] - 2*x^2 + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - 2*x^2*#1 - Sqrt[3 + 4*x^4]*(1 + #1)] + Log[2*x^2 +
Sqrt[3 + 4*x^4]]*#1 - Log[I*Sqrt[3] - 2*x^2 + 2*x*Sqrt[2*x^2 + Sqrt[3 + 4*x^4]] - 2*x^2*#1 - Sqrt[3 + 4*x^4]*
(1 + #1)]*#1)/((4*I)*Sqrt[3]*c^2 + 6*d^2 + (12*I)*Sqrt[3]*c^2*#1 + 12*d^2*#1 + (6*I)*Sqrt[3]*c^2*#1^2 + 9*d^2*
#1^2 + 3*d^2*#1^3) & ])/Sqrt[4*c^4 + 3*d^4]

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {2 x^{2}+\sqrt {4 x^{4}+3}}}{\left (d x +c \right ) \sqrt {4 x^{4}+3}}\, dx\]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)/(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)), x)

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Fricas [F(-1)] Timed out
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)/(4*x^4+3)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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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 ) \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)/(4*x**4+3)**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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)/(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)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {2\,x^2+\sqrt {4\,x^4+3}}}{\sqrt {4\,x^4+3}\,\left (c+d\,x\right )} \,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)),x)

[Out]

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

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