3.38 \(\int \sin ^2(e+f x) (a+b \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=46 \[ -\frac {(a-b) \sin (e+f x) \cos (e+f x)}{2 f}+\frac {1}{2} x (a-3 b)+\frac {b \tan (e+f x)}{f} \]

[Out]

1/2*(a-3*b)*x-1/2*(a-b)*cos(f*x+e)*sin(f*x+e)/f+b*tan(f*x+e)/f

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Rubi [A]  time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3663, 455, 388, 203} \[ -\frac {(a-b) \sin (e+f x) \cos (e+f x)}{2 f}+\frac {1}{2} x (a-3 b)+\frac {b \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - 3*b)*x)/2 - ((a - b)*Cos[e + f*x]*Sin[e + f*x])/(2*f) + (b*Tan[e + f*x])/f

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a-b) \cos (e+f x) \sin (e+f x)}{2 f}-\frac {\operatorname {Subst}\left (\int \frac {-a+b-2 b x^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(a-b) \cos (e+f x) \sin (e+f x)}{2 f}+\frac {b \tan (e+f x)}{f}+\frac {(a-3 b) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {1}{2} (a-3 b) x-\frac {(a-b) \cos (e+f x) \sin (e+f x)}{2 f}+\frac {b \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 43, normalized size = 0.93 \[ \frac {2 (a-3 b) (e+f x)+(b-a) \sin (2 (e+f x))+4 b \tan (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a - 3*b)*(e + f*x) + (-a + b)*Sin[2*(e + f*x)] + 4*b*Tan[e + f*x])/(4*f)

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fricas [A]  time = 0.42, size = 54, normalized size = 1.17 \[ \frac {{\left (a - 3 \, b\right )} f x \cos \left (f x + e\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right )}{2 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a - 3*b)*f*x*cos(f*x + e) - ((a - b)*cos(f*x + e)^2 - 2*b)*sin(f*x + e))/(f*cos(f*x + e))

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giac [B]  time = 1.97, size = 395, normalized size = 8.59 \[ \frac {a f x \tan \left (f x\right )^{3} \tan \relax (e)^{3} - 3 \, b f x \tan \left (f x\right )^{3} \tan \relax (e)^{3} + a f x \tan \left (f x\right )^{3} \tan \relax (e) - 3 \, b f x \tan \left (f x\right )^{3} \tan \relax (e) - a f x \tan \left (f x\right )^{2} \tan \relax (e)^{2} + 3 \, b f x \tan \left (f x\right )^{2} \tan \relax (e)^{2} + a f x \tan \left (f x\right ) \tan \relax (e)^{3} - 3 \, b f x \tan \left (f x\right ) \tan \relax (e)^{3} + a \tan \left (f x\right )^{3} \tan \relax (e)^{2} - 3 \, b \tan \left (f x\right )^{3} \tan \relax (e)^{2} + a \tan \left (f x\right )^{2} \tan \relax (e)^{3} - 3 \, b \tan \left (f x\right )^{2} \tan \relax (e)^{3} - a f x \tan \left (f x\right )^{2} + 3 \, b f x \tan \left (f x\right )^{2} + a f x \tan \left (f x\right ) \tan \relax (e) - 3 \, b f x \tan \left (f x\right ) \tan \relax (e) - a f x \tan \relax (e)^{2} + 3 \, b f x \tan \relax (e)^{2} - 2 \, b \tan \left (f x\right )^{3} - 2 \, a \tan \left (f x\right )^{2} \tan \relax (e) - 2 \, a \tan \left (f x\right ) \tan \relax (e)^{2} - 2 \, b \tan \relax (e)^{3} - a f x + 3 \, b f x + a \tan \left (f x\right ) - 3 \, b \tan \left (f x\right ) + a \tan \relax (e) - 3 \, b \tan \relax (e)}{2 \, {\left (f \tan \left (f x\right )^{3} \tan \relax (e)^{3} + f \tan \left (f x\right )^{3} \tan \relax (e) - f \tan \left (f x\right )^{2} \tan \relax (e)^{2} + f \tan \left (f x\right ) \tan \relax (e)^{3} - f \tan \left (f x\right )^{2} + f \tan \left (f x\right ) \tan \relax (e) - f \tan \relax (e)^{2} - f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*f*x*tan(f*x)^3*tan(e)^3 - 3*b*f*x*tan(f*x)^3*tan(e)^3 + a*f*x*tan(f*x)^3*tan(e) - 3*b*f*x*tan(f*x)^3*ta
n(e) - a*f*x*tan(f*x)^2*tan(e)^2 + 3*b*f*x*tan(f*x)^2*tan(e)^2 + a*f*x*tan(f*x)*tan(e)^3 - 3*b*f*x*tan(f*x)*ta
n(e)^3 + a*tan(f*x)^3*tan(e)^2 - 3*b*tan(f*x)^3*tan(e)^2 + a*tan(f*x)^2*tan(e)^3 - 3*b*tan(f*x)^2*tan(e)^3 - a
*f*x*tan(f*x)^2 + 3*b*f*x*tan(f*x)^2 + a*f*x*tan(f*x)*tan(e) - 3*b*f*x*tan(f*x)*tan(e) - a*f*x*tan(e)^2 + 3*b*
f*x*tan(e)^2 - 2*b*tan(f*x)^3 - 2*a*tan(f*x)^2*tan(e) - 2*a*tan(f*x)*tan(e)^2 - 2*b*tan(e)^3 - a*f*x + 3*b*f*x
+ a*tan(f*x) - 3*b*tan(f*x) + a*tan(e) - 3*b*tan(e))/(f*tan(f*x)^3*tan(e)^3 + f*tan(f*x)^3*tan(e) - f*tan(f*x
)^2*tan(e)^2 + f*tan(f*x)*tan(e)^3 - f*tan(f*x)^2 + f*tan(f*x)*tan(e) - f*tan(e)^2 - f)

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maple [A]  time = 0.45, size = 81, normalized size = 1.76 \[ \frac {a \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+b \left (\frac {\sin ^{5}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(a*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+b*(sin(f*x+e)^5/cos(f*x+e)+(sin(f*x+e)^3+3/2*sin(f*x+e))*cos
(f*x+e)-3/2*f*x-3/2*e))

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maxima [A]  time = 1.00, size = 51, normalized size = 1.11 \[ \frac {{\left (f x + e\right )} {\left (a - 3 \, b\right )} + 2 \, b \tan \left (f x + e\right ) - \frac {{\left (a - b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((f*x + e)*(a - 3*b) + 2*b*tan(f*x + e) - (a - b)*tan(f*x + e)/(tan(f*x + e)^2 + 1))/f

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mupad [B]  time = 11.32, size = 41, normalized size = 0.89 \[ \frac {b\,\mathrm {tan}\left (e+f\,x\right )-\sin \left (2\,e+2\,f\,x\right )\,\left (\frac {a}{4}-\frac {b}{4}\right )+f\,x\,\left (\frac {a}{2}-\frac {3\,b}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*tan(e + f*x) - sin(2*e + 2*f*x)*(a/4 - b/4) + f*x*(a/2 - (3*b)/2))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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