3.24 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=64 \[ \frac{a^2 \tan ^3(c+d x)}{5 d}+\frac{3 a^2 \tan (c+d x)}{5 d}+\frac{2 \sec ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{5 d} \]

[Out]

(2*Sec[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(5*d) + (3*a^2*Tan[c + d*x])/(5*d) + (a^2*Tan[c + d*x]^3)/(5*d)

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Rubi [A]  time = 0.0562102, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2676, 3767} \[ \frac{a^2 \tan ^3(c+d x)}{5 d}+\frac{3 a^2 \tan (c+d x)}{5 d}+\frac{2 \sec ^5(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2,x]

[Out]

(2*Sec[c + d*x]^5*(a^2 + a^2*Sin[c + d*x]))/(5*d) + (3*a^2*Tan[c + d*x])/(5*d) + (a^2*Tan[c + d*x]^3)/(5*d)

Rule 2676

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{2 \sec ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{5 d}+\frac{1}{5} \left (3 a^2\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{2 \sec ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{5 d}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{2 \sec ^5(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{5 d}+\frac{3 a^2 \tan (c+d x)}{5 d}+\frac{a^2 \tan ^3(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.0106957, size = 82, normalized size = 1.28 \[ \frac{2 a^2 \tan ^5(c+d x)}{5 d}+\frac{2 a^2 \sec ^5(c+d x)}{5 d}-\frac{a^2 \tan ^3(c+d x) \sec ^2(c+d x)}{d}+\frac{a^2 \tan (c+d x) \sec ^4(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^2,x]

[Out]

(2*a^2*Sec[c + d*x]^5)/(5*d) + (a^2*Sec[c + d*x]^4*Tan[c + d*x])/d - (a^2*Sec[c + d*x]^2*Tan[c + d*x]^3)/d + (
2*a^2*Tan[c + d*x]^5)/(5*d)

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Maple [A]  time = 0.103, size = 93, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{\frac{2\,{a}^{2}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{a}^{2} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)+2/5*a^2/cos(d*x+c)^5-a^2*(-8/15-1/5*se
c(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]  time = 0.975744, size = 104, normalized size = 1.62 \begin{align*} \frac{{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} +{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a^{2} + \frac{6 \, a^{2}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^2 + (3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*a^2
+ 6*a^2/cos(d*x + c)^5)/d

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Fricas [A]  time = 1.53434, size = 207, normalized size = 3.23 \begin{align*} -\frac{4 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} -{\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2}\right )} \sin \left (d x + c\right )}{5 \,{\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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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)**6*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.17407, size = 143, normalized size = 2.23 \begin{align*} -\frac{\frac{5 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} + \frac{35 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 90 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 120 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 70 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21 \, a^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{5}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/20*(5*a^2/(tan(1/2*d*x + 1/2*c) + 1) + (35*a^2*tan(1/2*d*x + 1/2*c)^4 - 90*a^2*tan(1/2*d*x + 1/2*c)^3 + 120
*a^2*tan(1/2*d*x + 1/2*c)^2 - 70*a^2*tan(1/2*d*x + 1/2*c) + 21*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^5)/d