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]

(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])

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Rubi [A]  time = 0.379546, antiderivative size = 411, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, 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 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]

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 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 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]

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 = 6.07386, 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)]])

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Maple [F]  time = 0.71, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}{\frac{1}{\sqrt{a+b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}}}\, dx \end{align*}

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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{2}}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (c + d x \right )}}{\sqrt{a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \end{align*}

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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{2}}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

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)