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]

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

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Rubi [A]  time = 0.19, antiderivative size = 135, normalized size of antiderivative = 1.00, 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 8

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

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Mathematica [A]  time = 0.54, size = 190, normalized size = 1.41 \[ \frac {\sec ^5(c+d x) \left (640 a^3 \sin (c+d x)+320 a^3 \sin (3 (c+d x))+64 a^3 \sin (5 (c+d x))+\left (110 b^3-270 a^2 b\right ) \cos (c+d x)-135 a^2 b \cos (3 (c+d x))-27 a^2 b \cos (5 (c+d x))+1152 a^2 b+960 a b^2 \sin (c+d x)-240 a b^2 \sin (3 (c+d x))-48 a b^2 \sin (5 (c+d x))-320 b^3 \cos (2 (c+d x))+55 b^3 \cos (3 (c+d x))+11 b^3 \cos (5 (c+d x))+64 b^3\right )}{1920 d} \]

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)

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fricas [A]  time = 0.43, size = 101, normalized size = 0.75 \[ -\frac {5 \, b^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{2} b - 3 \, b^{3} - {\left (2 \, {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, a^{3} + 9 \, a b^{2} + {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \]

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)

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giac [A]  time = 1.41, size = 243, normalized size = 1.80 \[ -\frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 58 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 90 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 20 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b - 2 \, b^{3}\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} d} \]

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)

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maple [A]  time = 0.32, size = 173, normalized size = 1.28 \[ \frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {3 a^{2} b}{5 \cos \left (d x +c \right )^{5}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{15 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{15}\right )}{d} \]

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)))

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maxima [A]  time = 0.33, size = 105, normalized size = 0.78 \[ \frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 3 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} - \frac {{\left (5 \, \cos \left (d x + c\right )^{2} - 3\right )} b^{3}}{\cos \left (d x + c\right )^{5}} + \frac {9 \, a^{2} b}{\cos \left (d x + c\right )^{5}}}{15 \, d} \]

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

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mupad [B]  time = 5.41, size = 119, normalized size = 0.88 \[ \frac {{\cos \left (c+d\,x\right )}^4\,\left (\frac {8\,a^3\,\sin \left (c+d\,x\right )}{15}-\frac {2\,a\,b^2\,\sin \left (c+d\,x\right )}{5}\right )-{\cos \left (c+d\,x\right )}^2\,\left (-\frac {4\,\sin \left (c+d\,x\right )\,a^3}{15}+\frac {\sin \left (c+d\,x\right )\,a\,b^2}{5}+\frac {b^3}{3}\right )+\frac {3\,a^2\,b}{5}+\frac {a^3\,\sin \left (c+d\,x\right )}{5}+\frac {b^3}{5}+\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{5}}{d\,{\cos \left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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