\(\int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [137]

Optimal result

Integrand size = 23, antiderivative size = 139 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {2} d}+\frac {a^2 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d} \]

[Out]

1/6*a*sec(d*x+c)^3*(a+a*sin(d*x+c))^(3/2)/d+1/5*sec(d*x+c)^5*(a+a*sin(d*x+c))^(5/2)/d-1/8*a^(5/2)*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)+1/4*a^2*sec(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2754, 2728, 212} \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{4 \sqrt {2} d}+\frac {a^2 \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}+\frac {\sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{5 d}+\frac {a \sec ^3(c+d x) (a \sin (c+d x)+a)^{3/2}}{6 d} \]

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

-1/4*(a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(Sqrt[2]*d) + (a^2*Sec[c + d
*x]*Sqrt[a + a*Sin[c + d*x]])/(4*d) + (a*Sec[c + d*x]^3*(a + a*Sin[c + d*x])^(3/2))/(6*d) + (Sec[c + d*x]^5*(a
+ a*Sin[c + d*x])^(5/2))/(5*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 ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}+\frac {1}{2} a \int \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = \frac {a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}+\frac {1}{4} a^2 \int \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = \frac {a^2 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}+\frac {1}{8} a^3 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {a^2 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}-\frac {a^3 \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 )}{4 d} \\ & = -\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{4 \sqrt {2} d}+\frac {a^2 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {a \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2}}{6 d}+\frac {\sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.37 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^5(c+d x) (a (1+\sin (c+d x)))^{5/2}}{5 d} \]

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.86

\[\frac {\left (1+\sin \left (d x +c \right )\right ) \left (-30 \left (\cos ^{2}\left (d x +c \right )\right ) a^{\frac {11}{2}}-80 \sin \left (d x +c \right ) a^{\frac {11}{2}}+104 a^{\frac {11}{2}}-15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}\right )}{120 a^{\frac {5}{2}} \left (\sin \left (d x +c \right )-1\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\]

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^(5/2),x)

[Out]

1/120*(1+sin(d*x+c))*(-30*cos(d*x+c)^2*a^(11/2)-80*sin(d*x+c)*a^(11/2)+104*a^(11/2)-15*2^(1/2)*arctanh(1/2*(a-
a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3*(a-a*sin(d*x+c))^(5/2))/a^(5/2)/(sin(d*x+c)-1)^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. 263 vs. \(2 (116) = 232\).

Time = 0.28 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.89 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {15 \, {\left (\sqrt {2} a^{2} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} + 40 \, a^{2} \sin \left (d x + c\right ) - 52 \, a^{2}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{240 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/240*(15*(sqrt(2)*a^2*cos(d*x + c)^3 + 2*sqrt(2)*a^2*cos(d*x + c)*sin(d*x + c) - 2*sqrt(2)*a^2*cos(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) + sqr
t(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*(15*a^2*cos(d*x + c)^2 + 40*a^2*sin(d*x + c) - 52*a^2)*sqrt(a*sin(d
*x + c) + a))/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)*sin(d*x + c) - 2*d*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**6*(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.80 \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} a^{\frac {5}{2}} {\left (\frac {2 \, {\left (15 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - 15 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 15 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{240 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/240*sqrt(2)*a^(5/2)*(2*(15*sin(-1/4*pi + 1/2*d*x + 1/2*c)^4 + 5*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 3)/sin(-
1/4*pi + 1/2*d*x + 1/2*c)^5 - 15*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 15*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. \[ \int \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^6} \,d x \]

[In]

int((a + a*sin(c + d*x))^(5/2)/cos(c + d*x)^6,x)

[Out]

int((a + a*sin(c + d*x))^(5/2)/cos(c + d*x)^6, x)