3.410 \(\int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=135 \[ \frac{2 a \left (4 a^2-3 b^2\right ) \tan (c+d x)}{15 d}+\frac{2 b \left (2 a^2-b^2\right ) \sec (c+d x)}{15 d}+\frac{2 \sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{15 d}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d} \]

[Out]

(2*b*(2*a^2 - b^2)*Sec[c + d*x])/(15*d) + (Sec[c + d*x]^5*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(5*d) +
(2*Sec[c + d*x]^3*(a + b*Sin[c + d*x])*(a*b + (2*a^2 - b^2)*Sin[c + d*x]))/(15*d) + (2*a*(4*a^2 - 3*b^2)*Tan[
c + d*x])/(15*d)

________________________________________________________________________________________

Rubi [A]  time = 0.19216, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2691, 2861, 2669, 3767, 8} \[ \frac{2 a \left (4 a^2-3 b^2\right ) \tan (c+d x)}{15 d}+\frac{2 b \left (2 a^2-b^2\right ) \sec (c+d x)}{15 d}+\frac{2 \sec ^3(c+d x) (a+b \sin (c+d x)) \left (\left (2 a^2-b^2\right ) \sin (c+d x)+a b\right )}{15 d}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^3,x]

[Out]

(2*b*(2*a^2 - b^2)*Sec[c + d*x])/(15*d) + (Sec[c + d*x]^5*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(5*d) +
(2*Sec[c + d*x]^3*(a + b*Sin[c + d*x])*(a*b + (2*a^2 - b^2)*Sin[c + d*x]))/(15*d) + (2*a*(4*a^2 - 3*b^2)*Tan[
c + d*x])/(15*d)

Rule 2691

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

Rule 2861

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

Rule 2669

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

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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d}-\frac{1}{5} \int \sec ^4(c+d x) (a+b \sin (c+d x)) \left (-4 a^2+2 b^2-2 a b \sin (c+d x)\right ) \, dx\\ &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d}+\frac{2 \sec ^3(c+d x) (a+b \sin (c+d x)) \left (a b+\left (2 a^2-b^2\right ) \sin (c+d x)\right )}{15 d}+\frac{1}{15} \int \sec ^2(c+d x) \left (2 a \left (4 a^2-3 b^2\right )+2 b \left (2 a^2-b^2\right ) \sin (c+d x)\right ) \, dx\\ &=\frac{2 b \left (2 a^2-b^2\right ) \sec (c+d x)}{15 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d}+\frac{2 \sec ^3(c+d x) (a+b \sin (c+d x)) \left (a b+\left (2 a^2-b^2\right ) \sin (c+d x)\right )}{15 d}+\frac{1}{15} \left (2 a \left (4 a^2-3 b^2\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{2 b \left (2 a^2-b^2\right ) \sec (c+d x)}{15 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d}+\frac{2 \sec ^3(c+d x) (a+b \sin (c+d x)) \left (a b+\left (2 a^2-b^2\right ) \sin (c+d x)\right )}{15 d}-\frac{\left (2 a \left (4 a^2-3 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{2 b \left (2 a^2-b^2\right ) \sec (c+d x)}{15 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d}+\frac{2 \sec ^3(c+d x) (a+b \sin (c+d x)) \left (a b+\left (2 a^2-b^2\right ) \sin (c+d x)\right )}{15 d}+\frac{2 a \left (4 a^2-3 b^2\right ) \tan (c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.623113, size = 190, normalized size = 1.41 \[ \frac{\sec ^5(c+d x) \left (\left (110 b^3-270 a^2 b\right ) \cos (c+d x)-135 a^2 b \cos (3 (c+d x))-27 a^2 b \cos (5 (c+d x))+1152 a^2 b+640 a^3 \sin (c+d x)+320 a^3 \sin (3 (c+d x))+64 a^3 \sin (5 (c+d x))+960 a b^2 \sin (c+d x)-240 a b^2 \sin (3 (c+d x))-48 a b^2 \sin (5 (c+d x))-320 b^3 \cos (2 (c+d x))+55 b^3 \cos (3 (c+d x))+11 b^3 \cos (5 (c+d x))+64 b^3\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^5*(1152*a^2*b + 64*b^3 + (-270*a^2*b + 110*b^3)*Cos[c + d*x] - 320*b^3*Cos[2*(c + d*x)] - 135*a^
2*b*Cos[3*(c + d*x)] + 55*b^3*Cos[3*(c + d*x)] - 27*a^2*b*Cos[5*(c + d*x)] + 11*b^3*Cos[5*(c + d*x)] + 640*a^3
*Sin[c + d*x] + 960*a*b^2*Sin[c + d*x] + 320*a^3*Sin[3*(c + d*x)] - 240*a*b^2*Sin[3*(c + d*x)] + 64*a^3*Sin[5*
(c + d*x)] - 48*a*b^2*Sin[5*(c + d*x)]))/(1920*d)

________________________________________________________________________________________

Maple [A]  time = 0.085, size = 173, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( -{a}^{3} \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 ) +{\frac{3\,{a}^{2}b}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+3\,a{b}^{2} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +{b}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{15\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{15}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+b*sin(d*x+c))^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 0.972075, size = 142, normalized size = 1.05 \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^{3} + 3 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} - \frac{{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}} + \frac{9 \, a^{2} b}{\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+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/15*((3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 3*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*a
*b^2 - (5*cos(d*x + c)^2 - 3)*b^3/cos(d*x + c)^5 + 9*a^2*b/cos(d*x + c)^5)/d

________________________________________________________________________________________

Fricas [A]  time = 2.39549, size = 232, normalized size = 1.72 \begin{align*} -\frac{5 \, b^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{2} b - 3 \, b^{3} -{\left (2 \,{\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{3} + 9 \, a b^{2} +{\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/15*(5*b^3*cos(d*x + c)^2 - 9*a^2*b - 3*b^3 - (2*(4*a^3 - 3*a*b^2)*cos(d*x + c)^4 + 3*a^3 + 9*a*b^2 + (4*a^3
- 3*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

________________________________________________________________________________________

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+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.10546, size = 328, normalized size = 2.43 \begin{align*} -\frac{2 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 45 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} - 20 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 60 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 30 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 58 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 90 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 10 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 20 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 60 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, a^{2} b - 2 \, b^{3}\right )}}{15 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-2/15*(15*a^3*tan(1/2*d*x + 1/2*c)^9 + 45*a^2*b*tan(1/2*d*x + 1/2*c)^8 - 20*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*a*
b^2*tan(1/2*d*x + 1/2*c)^7 + 30*b^3*tan(1/2*d*x + 1/2*c)^6 + 58*a^3*tan(1/2*d*x + 1/2*c)^5 + 24*a*b^2*tan(1/2*
d*x + 1/2*c)^5 + 90*a^2*b*tan(1/2*d*x + 1/2*c)^4 + 10*b^3*tan(1/2*d*x + 1/2*c)^4 - 20*a^3*tan(1/2*d*x + 1/2*c)
^3 + 60*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 10*b^3*tan(1/2*d*x + 1/2*c)^2 + 15*a^3*tan(1/2*d*x + 1/2*c) + 9*a^2*b -
2*b^3)/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*d)