\(\int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [155]

Optimal result

Integrand size = 23, antiderivative size = 233 \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {11 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{64 \sqrt {2} d}-\frac {11 a^5 \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}+\frac {11 a^4 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {11 a^3 \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{120 d}+\frac {11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d} \]

[Out]

-11/64*a^5*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)+11/140*a^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^(3/2)/d+11/126*a*sec(d
*x+c)^7*(a+a*sin(d*x+c))^(5/2)/d+1/9*sec(d*x+c)^9*(a+a*sin(d*x+c))^(7/2)/d-11/128*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)+11/48*a^4*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+11/120*a^3*
sec(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2754, 2766, 2729, 2728, 212} \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {11 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{64 \sqrt {2} d}-\frac {11 a^5 \cos (c+d x)}{64 d (a \sin (c+d x)+a)^{3/2}}+\frac {11 a^4 \sec (c+d x)}{48 d \sqrt {a \sin (c+d x)+a}}+\frac {11 a^3 \sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{120 d}+\frac {11 a^2 \sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{140 d}+\frac {\sec ^9(c+d x) (a \sin (c+d x)+a)^{7/2}}{9 d}+\frac {11 a \sec ^7(c+d x) (a \sin (c+d x)+a)^{5/2}}{126 d} \]

[In]

Int[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-11*a^(7/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(64*Sqrt[2]*d) - (11*a^5*Cos[
c + d*x])/(64*d*(a + a*Sin[c + d*x])^(3/2)) + (11*a^4*Sec[c + d*x])/(48*d*Sqrt[a + a*Sin[c + d*x]]) + (11*a^3*
Sec[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(120*d) + (11*a^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(140*d)
+ (11*a*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^(5/2))/(126*d) + (Sec[c + d*x]^9*(a + a*Sin[c + d*x])^(7/2))/(9*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 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

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]

Rule 2766

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*((
g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0
] && LtQ[p, -1] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac {1}{18} (11 a) \int \sec ^8(c+d x) (a+a \sin (c+d x))^{5/2} \, dx \\ & = \frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac {1}{28} \left (11 a^2\right ) \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = \frac {11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac {1}{40} \left (11 a^3\right ) \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {11 a^3 \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{120 d}+\frac {11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac {1}{48} \left (11 a^4\right ) \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {11 a^4 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {11 a^3 \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{120 d}+\frac {11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac {1}{32} \left (11 a^5\right ) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {11 a^5 \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}+\frac {11 a^4 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {11 a^3 \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{120 d}+\frac {11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac {1}{128} \left (11 a^4\right ) \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {11 a^5 \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}+\frac {11 a^4 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {11 a^3 \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{120 d}+\frac {11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}-\frac {\left (11 a^4\right ) \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 )}{64 d} \\ & = -\frac {11 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{64 \sqrt {2} d}-\frac {11 a^5 \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}+\frac {11 a^4 \sec (c+d x)}{48 d \sqrt {a+a \sin (c+d x)}}+\frac {11 a^3 \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{120 d}+\frac {11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac {11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac {\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.80 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.27 \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {a^3 \operatorname {Hypergeometric2F1}\left (-\frac {9}{2},2,-\frac {7}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^9(c+d x) (1+\sin (c+d x))^4 \sqrt {a (1+\sin (c+d x))}}{18 d} \]

[In]

Integrate[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(a^3*Hypergeometric2F1[-9/2, 2, -7/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^9*(1 + Sin[c + d*x])^4*Sqrt[a*(1 + Si
n[c + d*x])])/(18*d)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.89

\[-\frac {-6930 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right ) a^{\frac {11}{2}}+25410 \left (\cos ^{4}\left (d x +c \right )\right ) a^{\frac {11}{2}}+42504 \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) a^{\frac {11}{2}}+3465 \left (a -a \sin \left (d x +c \right )\right )^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right ) a -50424 \left (\cos ^{2}\left (d x +c \right )\right ) a^{\frac {11}{2}}+3465 \left (a -a \sin \left (d x +c \right )\right )^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -12320 \sin \left (d x +c \right ) a^{\frac {11}{2}}+7840 a^{\frac {11}{2}}}{40320 a^{\frac {3}{2}} \left (\sin \left (d x +c \right )-1\right )^{4} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\]

[In]

int(sec(d*x+c)^10*(a+a*sin(d*x+c))^(7/2),x)

[Out]

-1/40320/a^(3/2)*(-6930*sin(d*x+c)*cos(d*x+c)^4*a^(11/2)+25410*cos(d*x+c)^4*a^(11/2)+42504*sin(d*x+c)*cos(d*x+
c)^2*a^(11/2)+3465*(a-a*sin(d*x+c))^(9/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+
c)*a-50424*cos(d*x+c)^2*a^(11/2)+3465*(a-a*sin(d*x+c))^(9/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2
)/a^(1/2))*a-12320*sin(d*x+c)*a^(11/2)+7840*a^(11/2))/(sin(d*x+c)-1)^4/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.48 \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {3465 \, {\left (3 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{5} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3} - {\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{5} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (12705 \, a^{3} \cos \left (d x + c\right )^{4} - 25212 \, a^{3} \cos \left (d x + c\right )^{2} + 3920 \, a^{3} - 77 \, {\left (45 \, a^{3} \cos \left (d x + c\right )^{4} - 276 \, a^{3} \cos \left (d x + c\right )^{2} + 80 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{80640 \, {\left (3 \, d \cos \left (d x + c\right )^{5} - 4 \, d \cos \left (d x + c\right )^{3} - {\left (d \cos \left (d x + c\right )^{5} - 4 \, d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^10*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/80640*(3465*(3*sqrt(2)*a^3*cos(d*x + c)^5 - 4*sqrt(2)*a^3*cos(d*x + c)^3 - (sqrt(2)*a^3*cos(d*x + c)^5 - 4*s
qrt(2)*a^3*cos(d*x + c)^3)*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*(12705*a^3*cos(d*x + c)^4
- 25212*a^3*cos(d*x + c)^2 + 3920*a^3 - 77*(45*a^3*cos(d*x + c)^4 - 276*a^3*cos(d*x + c)^2 + 80*a^3)*sin(d*x +
c))*sqrt(a*sin(d*x + c) + a))/(3*d*cos(d*x + c)^5 - 4*d*cos(d*x + c)^3 - (d*cos(d*x + c)^5 - 4*d*cos(d*x + c)
^3)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**10*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^10*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.75 \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {\sqrt {2} a^{\frac {7}{2}} {\left (\frac {630 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {4 \, {\left (1575 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 420 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 189 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 90 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9}} - 3465 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 3465 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{80640 \, d} \]

[In]

integrate(sec(d*x+c)^10*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

-1/80640*sqrt(2)*a^(7/2)*(630*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1) + 4*(1575*
sin(-1/4*pi + 1/2*d*x + 1/2*c)^8 + 420*sin(-1/4*pi + 1/2*d*x + 1/2*c)^6 + 189*sin(-1/4*pi + 1/2*d*x + 1/2*c)^4
+ 90*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 35)/sin(-1/4*pi + 1/2*d*x + 1/2*c)^9 - 3465*log(sin(-1/4*pi + 1/2*d*x
+ 1/2*c) + 1) + 3465*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. \[ \int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^{10}} \,d x \]

[In]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^10,x)

[Out]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^10, x)