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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 = 0.01, size = 82, normalized size = 1.28 \[ \frac {2 a^2 \tan ^5(c+d x)}{5 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{d}-\frac {a^2 \tan ^3(c+d x) \sec ^2(c+d x)}{d} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 85, normalized size = 1.33 \[ -\frac {4 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2}\right )} \sin \left (d x + c\right )}{5 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \]

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))

________________________________________________________________________________________

giac [A]  time = 0.65, size = 106, normalized size = 1.66 \[ -\frac {\frac {5 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {35 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 90 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 70 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}}}{20 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.22, size = 93, normalized size = 1.45 \[ \frac {a^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+\frac {2 a^{2}}{5 \cos \left (d x +c \right )^{5}}-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d} \]

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))

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 77, normalized size = 1.20 \[ \frac {{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} + {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a^{2} + \frac {6 \, a^{2}}{\cos \left (d x + c\right )^{5}}}{15 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 4.63, size = 156, normalized size = 2.44 \[ \frac {2\,a^2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+10\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-10\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}{5\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*cos(c/2 + (d*x)/2)*(2*cos(c/2 + (d*x)/2)^5 + 5*sin(c/2 + (d*x)/2)^5 - 10*cos(c/2 + (d*x)/2)*sin(c/2 + (
d*x)/2)^4 - 3*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2) + 10*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^3))/(5*d*(c
os(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^5*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2)))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________