3.561 \(\int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

Optimal. Leaf size=411 \[ \frac {\sin (c+d x) \cos (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{d \sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )}+\frac {\sqrt [4]{a} \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 d (a+b)^{3/4} \sqrt {a+b \sin ^4(c+d x)}}-\frac {\sqrt [4]{a} \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{d (a+b)^{3/4} \sqrt {a+b \sin ^4(c+d x)}} \]

[Out]

-a^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))^2)^(1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x
+c)/a^(1/4)))*EllipticE(sin(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4))),1/2*(2-2*a^(1/2)/(a+b)^(1/2))^(1/2))*((a
+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)^2)^(1/2)*(a^(1/2)+(a+b)^(1/2)*tan(d*x
+c)^2)/(a+b)^(3/4)/d/(a+b*sin(d*x+c)^4)^(1/2)+1/2*a^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^
(1/4)))^2)^(1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticF(sin(2*arctan((a+b)^(1/4)*tan(d*x+c)/a
^(1/4))),1/2*(2-2*a^(1/2)/(a+b)^(1/2))^(1/2))*((a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*ta
n(d*x+c)^2)^2)^(1/2)*(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)/(a+b)^(3/4)/d/(a+b*sin(d*x+c)^4)^(1/2)+cos(d*x+c)*sin(
d*x+c)*(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)/d/(a+b)^(1/2)/(a+b*sin(d*x+c)^4)^(1/2)/(a^(1/2)+(a+b)^(1/2)*tan
(d*x+c)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 411, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3232, 1139, 1103, 1195} \[ \frac {\sin (c+d x) \cos (c+d x) \left ((a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}{d \sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )}+\frac {\sqrt [4]{a} \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 d (a+b)^{3/4} \sqrt {a+b \sin ^4(c+d x)}}-\frac {\sqrt [4]{a} \cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{d (a+b)^{3/4} \sqrt {a+b \sin ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

(Cos[c + d*x]*Sin[c + d*x]*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))/(Sqrt[a + b]*d*Sqrt[a + b*Sin[c
+ d*x]^4]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)) - (a^(1/4)*Cos[c + d*x]^2*EllipticE[2*ArcTan[((a + b)^(1/4)*
Tan[c + d*x])/a^(1/4)], (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[(a + 2*a*Tan[
c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/((a + b)^(3/4)*d*Sqrt[a + b*Si
n[c + d*x]^4]) + (a^(1/4)*Cos[c + d*x]^2*EllipticF[2*ArcTan[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)], (1 - Sqrt[a
]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4
)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/(2*(a + b)^(3/4)*d*Sqrt[a + b*Sin[c + d*x]^4])

Rule 1103

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(
a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1139

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

Rule 1195

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

Rule 3232

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{ff
= FreeFactors[Tan[e + f*x], x]}, Dist[(ff*(a + b*Sin[e + f*x]^4)^p*(Sec[e + f*x]^2)^(2*p))/(f*Apart[a*(1 + Tan
[e + f*x]^2)^2 + b*Tan[e + f*x]^4]^p), Subst[Int[((d*ff*x)^m*ExpandToSum[a*(1 + ff^2*x^2)^2 + b*ff^4*x^4, x]^p
)/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {\tan ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=\frac {\left (\cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\\ &=\frac {\left (\sqrt {a} \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{\sqrt {a+b} d \sqrt {a+b \sin ^4(c+d x)}}-\frac {\left (\sqrt {a} \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {a+b} x^2}{\sqrt {a}}}{\sqrt {a+2 a x^2+(a+b) x^4}} \, dx,x,\tan (c+d x)\right )}{\sqrt {a+b} d \sqrt {a+b \sin ^4(c+d x)}}\\ &=\frac {\cos (c+d x) \sin (c+d x) \left (a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)\right )}{\sqrt {a+b} d \sqrt {a+b \sin ^4(c+d x)} \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )}-\frac {\sqrt [4]{a} \cos ^2(c+d x) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{(a+b)^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\sqrt [4]{a} \cos ^2(c+d x) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 (a+b)^{3/4} d \sqrt {a+b \sin ^4(c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.15, size = 291, normalized size = 0.71 \[ -\frac {2 i \sqrt {2} \sqrt {a} \cos ^2(c+d x) \sqrt {1+\left (1-\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \left (E\left (i \sinh ^{-1}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )-F\left (i \sinh ^{-1}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right )|\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )\right )}{d \left (\sqrt {a}+i \sqrt {b}\right ) \sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \sqrt {8 a-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))+3 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^2/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

((-2*I)*Sqrt[2]*Sqrt[a]*Cos[c + d*x]^2*(EllipticE[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], (Sqrt
[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])] - EllipticF[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], (Sq
rt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])])*Sqrt[1 + (1 - (I*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2]*Sqrt[1 + (1 + (I
*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2])/((Sqrt[a] + I*Sqrt[b])*Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*d*Sqrt[8*a + 3*b - 4*
b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]])

________________________________________________________________________________________

fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\tan \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(tan(d*x + c)^2/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(d*x + c)^2/sqrt(b*sin(d*x + c)^4 + a), x)

________________________________________________________________________________________

maple [F]  time = 1.54, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{2}\left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(d*x + c)^2/sqrt(b*sin(d*x + c)^4 + a), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(c + d*x)**2/sqrt(a + b*sin(c + d*x)**4), x)

________________________________________________________________________________________