\(\int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx\) [495]

Optimal result

Integrand size = 25, antiderivative size = 171 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=-\frac {2 \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f} \]

[Out]

-2*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e)^2)^(1/2)/f/(1+b*sin(f*x+
e)^2/a)^(1/2)+a*EllipticF(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/
f/(a+b*sin(f*x+e)^2)^(1/2)+(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3275, 478, 538, 437, 435, 432, 430} \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {a \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\tan (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f} \]

[In]

Int[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x]^2,x]

[Out]

(-2*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(f*S
qrt[1 + (b*Sin[e + f*x]^2)/a]) + (a*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*
Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(f*Sqrt[a + b*Sin[e + f*x]^2]) + (Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/f

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]
, Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

Rule 3275

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff^(m + 1)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), Subst[Int[x^m*((a + b*ff^2*
x^2)^p/(1 - ff^2*x^2)^((m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^2 \sqrt {a+b x^2}}{\left (1-x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {a+2 b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}-\frac {\left (2 \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f}+\frac {\left (a \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}-\frac {\left (2 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (a \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a+b \sin ^2(e+f x)}} \\ & = -\frac {2 \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {-2 \sqrt {2} a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+(2 a+b-b \cos (2 (e+f x))) \tan (e+f x)}{\sqrt {2} f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[In]

Integrate[Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x]^2,x]

[Out]

(-2*Sqrt[2]*a*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + Sqrt[2]*a*Sqrt[(2*a + b - b*
Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + (2*a + b - b*Cos[2*(e + f*x)])*Tan[e + f*x])/(Sqrt[2]*f*Sqrt
[2*a + b - b*Cos[2*(e + f*x)]])

Maple [A] (verified)

Time = 3.76 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, a E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) b +a \sin \left (f x +e \right )+b \sin \left (f x +e \right )\right )}{\sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(222\)

[In]

int((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x,method=_RETURNVERBOSE)

[Out]

(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF
(sin(f*x+e),(-1/a*b)^(1/2))-2*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*a*EllipticE(sin(f*x+e),(-
1/a*b)^(1/2))-cos(f*x+e)^2*sin(f*x+e)*b+a*sin(f*x+e)+b*sin(f*x+e))/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(
f*x+e)))^(1/2)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

Fricas [F]

\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{2} \,d x } \]

[In]

integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="fricas")

[Out]

integral(sqrt(-b*cos(f*x + e)^2 + a + b)*tan(f*x + e)^2, x)

Sympy [F]

\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx \]

[In]

integrate((a+b*sin(f*x+e)**2)**(1/2)*tan(f*x+e)**2,x)

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*tan(e + f*x)**2, x)

Maxima [F]

\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{2} \,d x } \]

[In]

integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*tan(f*x + e)^2, x)

Giac [F]

\[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{2} \,d x } \]

[In]

integrate((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="giac")

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*tan(f*x + e)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+b \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]

[In]

int(tan(e + f*x)^2*(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(tan(e + f*x)^2*(a + b*sin(e + f*x)^2)^(1/2), x)