3.806 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=63 \[ \frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{5 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+a^2 x \]

[Out]

a^2*x - (5*a^2*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x])/(3*d*(a - a*Sin[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.206785, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2869, 2758, 2735, 2648} \[ \frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{5 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]

[Out]

a^2*x - (5*a^2*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x])/(3*d*(a - a*Sin[c + d*x])^2)

Rule 2869

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]

Rule 2758

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*Cos[e + f*x]*(
a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b
*(2*m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx &=a^4 \int \frac{\sin ^2(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} a^2 \int \frac{-2 a-3 a \sin (c+d x)}{a-a \sin (c+d x)} \, dx\\ &=a^2 x+\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{1}{3} \left (5 a^3\right ) \int \frac{1}{a-a \sin (c+d x)} \, dx\\ &=a^2 x+\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{5 a^3 \cos (c+d x)}{3 d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0514261, size = 79, normalized size = 1.25 \[ \frac{2 a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan ^{-1}(\tan (c+d x))}{d}-\frac{a^2 \tan (c+d x)}{d}+\frac{2 a^2 \sec ^3(c+d x)}{3 d}-\frac{2 a^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]

[Out]

(a^2*ArcTan[Tan[c + d*x]])/d - (2*a^2*Sec[c + d*x])/d + (2*a^2*Sec[c + d*x]^3)/(3*d) - (a^2*Tan[c + d*x])/d +
(2*a^2*Tan[c + d*x]^3)/(3*d)

________________________________________________________________________________________

Maple [A]  time = 0.074, size = 114, normalized size = 1.8 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-\tan \left ( dx+c \right ) +dx+c \right ) +2\,{a}^{2} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}-1/3\, \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c)+2*a^2*(1/3*sin(d*x+c)^4/cos(d*x+c)^3-1/3*sin(d*x+c)^4/cos(d*x+c)-
1/3*(2+sin(d*x+c)^2)*cos(d*x+c))+1/3*a^2*sin(d*x+c)^3/cos(d*x+c)^3)

________________________________________________________________________________________

Maxima [A]  time = 1.65301, size = 96, normalized size = 1.52 \begin{align*} \frac{a^{2} \tan \left (d x + c\right )^{3} +{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - \frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/3*(a^2*tan(d*x + c)^3 + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^2 - 2*(3*cos(d*x + c)^2 - 1)*a^2/c
os(d*x + c)^3)/d

________________________________________________________________________________________

Fricas [B]  time = 1.41047, size = 329, normalized size = 5.22 \begin{align*} -\frac{6 \, a^{2} d x -{\left (3 \, a^{2} d x + 5 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} +{\left (3 \, a^{2} d x - 4 \, a^{2}\right )} \cos \left (d x + c\right ) -{\left (6 \, a^{2} d x - a^{2} +{\left (3 \, a^{2} d x - 5 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) +{\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(6*a^2*d*x - (3*a^2*d*x + 5*a^2)*cos(d*x + c)^2 + a^2 + (3*a^2*d*x - 4*a^2)*cos(d*x + c) - (6*a^2*d*x - a
^2 + (3*a^2*d*x - 5*a^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*
d)*sin(d*x + c) - 2*d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)**2*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.21985, size = 90, normalized size = 1.43 \begin{align*} \frac{3 \,{\left (d x + c\right )} a^{2} + \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*a^2 + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*tan(1/2*d*x + 1/2*c) + 4*a^2)/(tan(1/2*d*x + 1/
2*c) - 1)^3)/d