\(\int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [153]

Optimal result

Integrand size = 23, antiderivative size = 171 \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {2} d}+\frac {a^3 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}+\frac {a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac {a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac {\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d} \]

[Out]

1/12*a^2*sec(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d+1/10*a*sec(d*x+c)^5*(a+a*sin(d*x+c))^(5/2)/d+1/7*sec(d*x+c)^7*(
a+a*sin(d*x+c))^(7/2)/d-1/16*a^(7/2)*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/d*2^(1/2)+
1/8*a^3*sec(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2754, 2728, 212} \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{8 \sqrt {2} d}+\frac {a^3 \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{8 d}+\frac {a^2 \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{12 d}+\frac {\sec ^7(c+d x) (a \sin (c+d x)+a)^{7/2}}{7 d}+\frac {a \sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{10 d} \]

[In]

Int[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

-1/8*(a^(7/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(Sqrt[2]*d) + (a^3*Sec[c + d
*x]*Sqrt[a + a*Sin[c + d*x]])/(8*d) + (a^2*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(12*d) + (a*Sec[c + d*x]
^5*(a + a*Sin[c + d*x])^(5/2))/(10*d) + (Sec[c + d*x]^7*(a + a*Sin[c + d*x])^(7/2))/(7*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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

Rule 2754

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac {1}{2} a \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx \\ & = \frac {a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac {\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac {1}{4} a^2 \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = \frac {a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac {a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac {\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac {1}{8} a^3 \int \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {a^3 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}+\frac {a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac {a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac {\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}+\frac {1}{16} a^4 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {a^3 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}+\frac {a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac {a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac {\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d}-\frac {a^4 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{8 d} \\ & = -\frac {a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {2} d}+\frac {a^3 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{8 d}+\frac {a^2 \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{12 d}+\frac {a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{10 d}+\frac {\sec ^7(c+d x) (a+a \sin (c+d x))^{7/2}}{7 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.47 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.37 \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {a^3 \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^7(c+d x) (1+\sin (c+d x))^3 \sqrt {a (1+\sin (c+d x))}}{7 d} \]

[In]

Integrate[Sec[c + d*x]^8*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(a^3*Hypergeometric2F1[-7/2, 1, -5/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^7*(1 + Sin[c + d*x])^3*Sqrt[a*(1 + Si
n[c + d*x])])/(7*d)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.81

\[\frac {\left (1+\sin \left (d x +c \right )\right ) \left (-210 a^{\frac {15}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+770 a^{\frac {15}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+1288 a^{\frac {15}{2}} \sin \left (d x +c \right )-1528 a^{\frac {15}{2}}+105 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4} \left (a -a \sin \left (d x +c \right )\right )^{\frac {7}{2}}\right )}{1680 a^{\frac {7}{2}} \left (\sin \left (d x +c \right )-1\right )^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\]

[In]

int(sec(d*x+c)^8*(a+a*sin(d*x+c))^(7/2),x)

[Out]

1/1680/a^(7/2)*(1+sin(d*x+c))/(sin(d*x+c)-1)^3*(-210*a^(15/2)*cos(d*x+c)^2*sin(d*x+c)+770*a^(15/2)*cos(d*x+c)^
2+1288*a^(15/2)*sin(d*x+c)-1528*a^(15/2)+105*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4*(
a-a*sin(d*x+c))^(7/2))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (144) = 288\).

Time = 0.30 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.82 \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {105 \, {\left (3 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right ) - {\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{3} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (385 \, a^{3} \cos \left (d x + c\right )^{2} - 764 \, a^{3} - 7 \, {\left (15 \, a^{3} \cos \left (d x + c\right )^{2} - 92 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3360 \, {\left (3 \, d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right ) - {\left (d \cos \left (d x + c\right )^{3} - 4 \, d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/3360*(105*(3*sqrt(2)*a^3*cos(d*x + c)^3 - 4*sqrt(2)*a^3*cos(d*x + c) - (sqrt(2)*a^3*cos(d*x + c)^3 - 4*sqrt(
2)*a^3*cos(d*x + c))*sin(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(a*sin(d*x + c) + a)*(sqrt(2)*cos(d*
x + c) - sqrt(2)*sin(d*x + c) + sqrt(2))*sqrt(a) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*
a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 4*(385*a^3*cos(d*x + c)^2 - 764*a^
3 - 7*(15*a^3*cos(d*x + c)^2 - 92*a^3)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(3*d*cos(d*x + c)^3 - 4*d*cos(d
*x + c) - (d*cos(d*x + c)^3 - 4*d*cos(d*x + c))*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**8*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.74 \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {\sqrt {2} a^{\frac {7}{2}} {\left (\frac {2 \, {\left (105 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 35 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 21 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7}} - 105 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 105 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{3360 \, d} \]

[In]

integrate(sec(d*x+c)^8*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

-1/3360*sqrt(2)*a^(7/2)*(2*(105*sin(-1/4*pi + 1/2*d*x + 1/2*c)^6 + 35*sin(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 21*si
n(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 15)/sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 105*log(sin(-1/4*pi + 1/2*d*x + 1/2*c)
+ 1) + 105*log(-sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d

Mupad [F(-1)]

Timed out. \[ \int \sec ^8(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^8} \,d x \]

[In]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^8,x)

[Out]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^8, x)