3.10 \(\int (a+b \sec ^2(e+f x)) \sin ^2(e+f x) \, dx\)

Optimal. Leaf size=42 \[ \frac{1}{2} x (a-2 b)-\frac{a \sin (e+f x) \cos (e+f x)}{2 f}+\frac{b \tan (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.0438048, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4132, 455, 388, 203} \[ \frac{1}{2} x (a-2 b)-\frac{a \sin (e+f x) \cos (e+f x)}{2 f}+\frac{b \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4132

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.096271, size = 54, normalized size = 1.29 \[ \frac{a (e+f x)}{2 f}-\frac{a \sin (2 (e+f x))}{4 f}-\frac{b \tan ^{-1}(\tan (e+f x))}{f}+\frac{b \tan (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.038, size = 46, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( a \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +b \left ( \tan \left ( fx+e \right ) -fx-e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.48669, size = 63, normalized size = 1.5 \begin{align*} \frac{{\left (f x + e\right )}{\left (a - 2 \, b\right )} + 2 \, b \tan \left (f x + e\right ) - \frac{a \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.477747, size = 123, normalized size = 2.93 \begin{align*} \frac{{\left (a - 2 \, b\right )} f x \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right )}{2 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sin ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15883, size = 69, normalized size = 1.64 \begin{align*} \frac{{\left (f x + e\right )}{\left (a - 2 \, b\right )} + 2 \, b \tan \left (f x + e\right ) - \frac{a \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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