\(\int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [113]

Optimal result

Integrand size = 23, antiderivative size = 197 \[ \int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {63 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{128 \sqrt {2} d}-\frac {63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d} \]

[Out]

-63/128*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-21/80*a^2*sec(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-63/256*arctanh(1
/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d*2^(1/2)+21/32*a*sec(d*x+c)/d/(a+a*sin(d*x+c))^
(1/2)+3/10*a*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+1/5*sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2754, 2766, 2760, 2729, 2728, 212} \[ \int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {63 a^2 \cos (c+d x)}{128 d (a \sin (c+d x)+a)^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a \sin (c+d x)+a)^{3/2}}-\frac {63 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{128 \sqrt {2} d}+\frac {\sec ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{5 d}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a \sin (c+d x)+a}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a \sin (c+d x)+a}} \]

[In]

Int[Sec[c + d*x]^6*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-63*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(128*Sqrt[2]*d) - (63*a^2*Cos
[c + d*x])/(128*d*(a + a*Sin[c + d*x])^(3/2)) - (21*a^2*Sec[c + d*x])/(80*d*(a + a*Sin[c + d*x])^(3/2)) + (21*
a*Sec[c + d*x])/(32*d*Sqrt[a + a*Sin[c + d*x]]) + (3*a*Sec[c + d*x]^3)/(10*d*Sqrt[a + a*Sin[c + d*x]]) + (Sec[
c + d*x]^5*Sqrt[a + a*Sin[c + d*x]])/(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 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 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1))), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 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 ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{10} (9 a) \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = \frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{20} \left (21 a^2\right ) \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{32} (21 a) \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{64} \left (63 a^2\right ) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx \\ & = -\frac {63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{256} (63 a) \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx \\ & = -\frac {63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}-\frac {(63 a) \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 )}{128 d} \\ & = -\frac {63 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{128 \sqrt {2} d}-\frac {63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.31 \[ \int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},3,-\frac {3}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^5(c+d x) (1+\sin (c+d x))^2 \sqrt {a (1+\sin (c+d x))}}{20 d} \]

[In]

Integrate[Sec[c + d*x]^6*Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(Hypergeometric2F1[-5/2, 3, -3/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^5*(1 + Sin[c + d*x])^2*Sqrt[a*(1 + Sin[c
+ d*x])])/(20*d)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.29

method	result	size
default	\(-\frac {-630 a^{\frac {9}{2}} \left (\cos ^{4}\left (d x +c \right )\right )-420 a^{\frac {9}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-315 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, a^{2} \left (\cos ^{2}\left (d x +c \right )\right )+84 a^{\frac {9}{2}} \left (\cos ^{2}\left (d x +c \right )\right )+630 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right ) a^{2}-288 a^{\frac {9}{2}} \sin \left (d x +c \right )+630 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+32 a^{\frac {9}{2}}}{1280 a^{\frac {7}{2}} \left (\sin \left (d x +c \right )-1\right )^{2} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(254\)

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/1280/a^(7/2)*(-630*a^(9/2)*cos(d*x+c)^4-420*a^(9/2)*cos(d*x+c)^2*sin(d*x+c)-315*arctanh(1/2*(a-a*sin(d*x+c)
)^(1/2)*2^(1/2)/a^(1/2))*(a-a*sin(d*x+c))^(5/2)*2^(1/2)*a^2*cos(d*x+c)^2+84*a^(9/2)*cos(d*x+c)^2+630*(a-a*sin(
d*x+c))^(5/2)*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^2-288*a^(9/2)*sin(d*x+c
)+630*(a-a*sin(d*x+c))^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+32*a^(9/2))/(sin(
d*x+c)-1)^2/(1+sin(d*x+c))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.07 \[ \int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {315 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{5} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 42 \, \cos \left (d x + c\right )^{2} + 6 \, {\left (35 \, \cos \left (d x + c\right )^{2} + 24\right )} \sin \left (d x + c\right ) - 16\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2560 \, d \cos \left (d x + c\right )^{5}} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2560*(315*sqrt(2)*sqrt(a)*cos(d*x + c)^5*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*(315*cos(d*x + c)^4 - 42*cos(d*x
+ c)^2 + 6*(35*cos(d*x + c)^2 + 24)*sin(d*x + c) - 16)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^5)

Sympy [F(-1)]

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

[In]

integrate(sec(d*x+c)**6*(a+a*sin(d*x+c))**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{6} \,d x } \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*sin(d*x + c) + a)*sec(d*x + c)^6, x)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.21 \[ \int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (315 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 315 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {10 \, {\left (15 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 17 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {16 \, {\left (30 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}\right )} \sqrt {a}}{2560 \, d} \]

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/2560*sqrt(2)*(315*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)) - 315*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)) - 10*(15*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*
c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 17*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))/
(sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2 - 16*(30*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x +
1/2*c)^4 + 5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 + sgn(cos(-1/4*pi + 1/2*d*x
+ 1/2*c)))/sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^6} \,d x \]

[In]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^6,x)

[Out]

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^6, x)