3.388 \(\int \frac{1}{\sqrt{a+b \tan ^4(c+d x)}} \, dx\)

Optimal. Leaf size=348 \[ -\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a+b \tan ^4(c+d x)}}\right )}{2 d \sqrt{a+b}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}} \]

[Out]

ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^4]]/(2*Sqrt[a + b]*d) - (b^(1/4)*EllipticF[2*ArcTan[
(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a]
+ Sqrt[b]*Tan[c + d*x]^2)^2])/(2*a^(1/4)*(Sqrt[a] - Sqrt[b])*d*Sqrt[a + b*Tan[c + d*x]^4]) + ((Sqrt[a] + Sqrt[
b])*EllipticPi[-(Sqrt[a] - Sqrt[b])^2/(4*Sqrt[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqr
t[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)^2])/(4*a^(1/4)*(
Sqrt[a] - Sqrt[b])*b^(1/4)*d*Sqrt[a + b*Tan[c + d*x]^4])

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Rubi [A]  time = 0.224285, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3661, 1217, 220, 1707} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a+b \tan ^4(c+d x)}}\right )}{2 d \sqrt{a+b}}-\frac{\sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Tan[c + d*x]^4],x]

[Out]

ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^4]]/(2*Sqrt[a + b]*d) - (b^(1/4)*EllipticF[2*ArcTan[
(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a]
+ Sqrt[b]*Tan[c + d*x]^2)^2])/(2*a^(1/4)*(Sqrt[a] - Sqrt[b])*d*Sqrt[a + b*Tan[c + d*x]^4]) + ((Sqrt[a] + Sqrt[
b])*EllipticPi[-(Sqrt[a] - Sqrt[b])^2/(4*Sqrt[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqr
t[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)^2])/(4*a^(1/4)*(
Sqrt[a] - Sqrt[b])*b^(1/4)*d*Sqrt[a + b*Tan[c + d*x]^4])

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 1217

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

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+b \tan ^4(c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\sqrt{a} \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{b} x^2}{\sqrt{a}}}{\left (1+x^2\right ) \sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{\left (\sqrt{a}-\sqrt{b}\right ) d}-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{\left (\sqrt{a}-\sqrt{b}\right ) d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a+b \tan ^4(c+d x)}}\right )}{2 \sqrt{a+b} d}-\frac{\sqrt [4]{b} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}}}{2 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{b}\right ) d \sqrt{a+b \tan ^4(c+d x)}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b} d \sqrt{a+b \tan ^4(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.360024, size = 106, normalized size = 0.3 \[ -\frac{i \sqrt{\frac{b \tan ^4(c+d x)}{a}+1} \Pi \left (-\frac{i \sqrt{a}}{\sqrt{b}};\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right )\right |-1\right )}{d \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b \tan ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*Tan[c + d*x]^4],x]

[Out]

((-I)*EllipticPi[((-I)*Sqrt[a])/Sqrt[b], I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], -1]*Sqrt[1 + (b*Ta
n[c + d*x]^4)/a])/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*d*Sqrt[a + b*Tan[c + d*x]^4])

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Maple [C]  time = 0.069, size = 123, normalized size = 0.4 \begin{align*}{\frac{1}{d}\sqrt{1-{i \left ( \tan \left ( dx+c \right ) \right ) ^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i \left ( \tan \left ( dx+c \right ) \right ) ^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticPi} \left ( \tan \left ( dx+c \right ) \sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},{i\sqrt{a}{\frac{1}{\sqrt{b}}}},{\sqrt{{-i\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}} \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( dx+c \right ) \right ) ^{4}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*tan(d*x+c)^4)^(1/2),x)

[Out]

1/d/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2
)/(a+b*tan(d*x+c)^4)^(1/2)*EllipticPi(tan(d*x+c)*(I/a^(1/2)*b^(1/2))^(1/2),I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/
2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*tan(d*x + c)^4 + a), x)

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Fricas [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/(a+tan(d*x+c)^4*b)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \tan ^{4}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+tan(d*x+c)**4*b)**(1/2),x)

[Out]

Integral(1/sqrt(a + b*tan(c + d*x)**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(b*tan(d*x + c)^4 + a), x)