3.6.17 \(\int \frac {\tan ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [517]

Optimal. Leaf size=109 \[ -\frac {\sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b) f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{(a+b) f} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 109, 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 = {3275, 482, 437, 435} \begin {gather*} \frac {\tan (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f (a+b)}-\frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f (a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((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])/((a +
b)*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a])) + (Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/((a + b)*f)

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 482

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 + 1)/(n*(b*c - a*d)*(p + 1))), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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

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Mathematica [A]
time = 0.27, size = 100, normalized size = 0.92 \begin {gather*} \frac {-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} (2 a+b-b \cos (2 (e+f x))) \tan (e+f x)}{2 (a+b) f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(221\) vs. \(2(102)=204\).
time = 17.38, size = 222, normalized size = 2.04

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.16, size = 730, normalized size = 6.70 \begin {gather*} \frac {2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b^{2} \sin \left (f x + e\right ) - {\left (2 i \, \sqrt {-b} b^{2} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right ) + {\left (2 i \, a b + i \, b^{2}\right )} \sqrt {-b} \cos \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) - {\left (-2 i \, \sqrt {-b} b^{2} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right ) + {\left (-2 i \, a b - i \, b^{2}\right )} \sqrt {-b} \cos \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) - 2 \, {\left (2 \, {\left (-i \, a b - i \, b^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right ) + {\left (2 i \, a^{2} + i \, a b\right )} \sqrt {-b} \cos \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) - 2 \, {\left (2 \, {\left (i \, a b + i \, b^{2}\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} \cos \left (f x + e\right ) + {\left (-2 i \, a^{2} - i \, a b\right )} \sqrt {-b} \cos \left (f x + e\right )\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{2 \, {\left (a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(-b*cos(f*x + e)^2 + a + b)*b^2*sin(f*x + e) - (2*I*sqrt(-b)*b^2*sqrt((a^2 + a*b)/b^2)*cos(f*x + e)
+ (2*I*a*b + I*b^2)*sqrt(-b)*cos(f*x + e))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sq
rt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b
+ b^2)*sqrt((a^2 + a*b)/b^2))/b^2) - (-2*I*sqrt(-b)*b^2*sqrt((a^2 + a*b)/b^2)*cos(f*x + e) + (-2*I*a*b - I*b^2
)*sqrt(-b)*cos(f*x + e))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 +
a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 +
a*b)/b^2))/b^2) - 2*(2*(-I*a*b - I*b^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2)*cos(f*x + e) + (2*I*a^2 + I*a*b)*sqrt(-
b)*cos(f*x + e))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^
2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2
))/b^2) - 2*(2*(I*a*b + I*b^2)*sqrt(-b)*sqrt((a^2 + a*b)/b^2)*cos(f*x + e) + (-2*I*a^2 - I*a*b)*sqrt(-b)*cos(f
*x + e))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a
+ b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2))
/((a*b^2 + b^3)*f*cos(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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