\(\int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx\) [388]

Optimal result

Integrand size = 16, antiderivative size = 348 \[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a+b \tan ^4(c+d x)}}\right )}{2 \sqrt {a+b} d}-\frac {\sqrt [4]{b} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right ) \left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+b \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )^2}}}{2 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) d \sqrt {a+b \tan ^4(c+d x)}}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2}{4 \sqrt {a} \sqrt {b}},2 \arctan \left (\frac {\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right ) \left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+b \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt {a}-\sqrt {b}\right ) \sqrt [4]{b} d \sqrt {a+b \tan ^4(c+d x)}} \]

[Out]

1/2*arctan((a+b)^(1/2)*tan(d*x+c)/(a+tan(d*x+c)^4*b)^(1/2))/d/(a+b)^(1/2)-1/2*b^(1/4)*(cos(2*arctan(b^(1/4)*ta
n(d*x+c)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*tan(d*x+c
)/a^(1/4))),1/2*2^(1/2))*((a+tan(d*x+c)^4*b)/(a^(1/2)+b^(1/2)*tan(d*x+c)^2)^2)^(1/2)*(a^(1/2)+b^(1/2)*tan(d*x+
c)^2)/a^(1/4)/d/(a^(1/2)-b^(1/2))/(a+tan(d*x+c)^4*b)^(1/2)+1/4*(cos(2*arctan(b^(1/4)*tan(d*x+c)/a^(1/4)))^2)^(
1/2)/cos(2*arctan(b^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticPi(sin(2*arctan(b^(1/4)*tan(d*x+c)/a^(1/4))),-1/4*(a^(1
/2)-b^(1/2))^2/a^(1/2)/b^(1/2),1/2*2^(1/2))*(a^(1/2)+b^(1/2))*((a+tan(d*x+c)^4*b)/(a^(1/2)+b^(1/2)*tan(d*x+c)^
2)^2)^(1/2)*(a^(1/2)+b^(1/2)*tan(d*x+c)^2)/a^(1/4)/b^(1/4)/d/(a^(1/2)-b^(1/2))/(a+tan(d*x+c)^4*b)^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3742, 1231, 226, 1721} \[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a+b \tan ^4(c+d x)}}\right )}{2 d \sqrt {a+b}}-\frac {\sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+b \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} d \left (\sqrt {a}-\sqrt {b}\right ) \sqrt {a+b \tan ^4(c+d x)}}+\frac {\left (\sqrt {a}+\sqrt {b}\right ) \left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+b \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {b} \tan ^2(c+d x)\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2}{4 \sqrt {a} \sqrt {b}},2 \arctan \left (\frac {\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt {a}-\sqrt {b}\right ) \sqrt {a+b \tan ^4(c+d x)}} \]

[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[-1/4*(Sqrt[a] - Sqrt[b])^2/(Sqrt[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(S
qrt[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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1231

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 1721

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))]/(4*d*e*A*q*Sqrt[a + c*x^4]))*El
lipticPi[Cancel[-(B*d - A*e)^2/(4*d*e*A*B)], 2*ArcTan[q*x], 1/2], 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]

Rule 3742

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])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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} \text {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} \text {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 {\arctan \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} \operatorname {EllipticF}\left (2 \arctan \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 ) \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {b}\right )^2}{4 \sqrt {a} \sqrt {b}},2 \arctan \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] (verified)

Result contains complex when optimal does not.

Time = 10.71 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=-\frac {i \operatorname {EllipticPi}\left (-\frac {i \sqrt {a}}{\sqrt {b}},i \text {arcsinh}\left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),-1\right ) \sqrt {1+\frac {b \tan ^4(c+d x)}{a}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {a+b \tan ^4(c+d x)}} \]

[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])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.57 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.35

method	result	size
derivativedivides	\(\frac {\sqrt {1-\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tan \left (d x +c \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, \frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{d \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +\tan \left (d x +c \right )^{4} b}}\)	\(123\)
default	\(\frac {\sqrt {1-\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, \tan \left (d x +c \right )^{2}}{\sqrt {a}}}\, \operatorname {EllipticPi}\left (\tan \left (d x +c \right ) \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, \frac {i \sqrt {a}}{\sqrt {b}}, \frac {\sqrt {-\frac {i \sqrt {b}}{\sqrt {a}}}}{\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}}\right )}{d \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {a +\tan \left (d x +c \right )^{4} b}}\)	\(123\)

[In]

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

[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+tan(d*x+c)^4*b)^(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))

Fricas [F]

\[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right )^{4} + a}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \tan ^{4}{\left (c + d x \right )}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right )^{4} + a}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right )^{4} + a}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \tan ^4(c+d x)}} \, dx=\int \frac {1}{\sqrt {b\,{\mathrm {tan}\left (c+d\,x\right )}^4+a}} \,d x \]

[In]

int(1/(a + b*tan(c + d*x)^4)^(1/2),x)

[Out]

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