\(\int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [150]

Optimal result

Integrand size = 23, antiderivative size = 106 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{8 d}+\frac {a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{2 d} \] Output:

-1/16*2^(1/2)*a^(7/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/ 
d-1/8*a^2*sec(d*x+c)^2*(a+a*sin(d*x+c))^(3/2)/d+1/2*a*sec(d*x+c)^4*(a+a*si 
n(d*x+c))^(5/2)/d
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {-\sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a (1+\sin (c+d x))}}{\sqrt {2} \sqrt {a}}\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+2 a^3 \sqrt {a (1+\sin (c+d x))} (3+\sin (c+d x))}{16 d (-1+\sin (c+d x))^2} \] Input:

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(7/2),x]
 

Output:

(-(Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])]*( 
Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + 2*a^3*Sqrt[a*(1 + Sin[c + d*x])] 
*(3 + Sin[c + d*x]))/(16*d*(-1 + Sin[c + d*x])^2)
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3155, 3042, 3154, 3042, 3146, 73, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec ^5(c+d x) (a \sin (c+d x)+a)^{7/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (c+d x)+a)^{7/2}}{\cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3155

\(\displaystyle \frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d}-\frac {1}{4} a^2 \int \sec ^3(c+d x) (\sin (c+d x) a+a)^{3/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d}-\frac {1}{4} a^2 \int \frac {(\sin (c+d x) a+a)^{3/2}}{\cos (c+d x)^3}dx\)

\(\Big \downarrow \) 3154

\(\displaystyle \frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d}-\frac {1}{4} a^2 \left (\frac {1}{4} a \int \sec (c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d}-\frac {1}{4} a^2 \left (\frac {1}{4} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\cos (c+d x)}dx+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\right )\)

\(\Big \downarrow \) 3146

\(\displaystyle \frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d}-\frac {1}{4} a^2 \left (\frac {a^2 \int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d}-\frac {1}{4} a^2 \left (\frac {a^2 \int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{2 d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{2 d}-\frac {1}{4} a^2 \left (\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{2 \sqrt {2} d}+\frac {\sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{2 d}\right )\)

Input:

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(7/2),x]
 

Output:

(a*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2))/(2*d) - (a^2*((a^(3/2)*ArcTa 
nh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]])/(2*Sqrt[2]*d) + (Sec[c + d*x]^2*(a + a 
*Sin[c + d*x])^(3/2))/(2*d)))/4
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3146

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x 
)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

rule 3154

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] + Simp[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 3155

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] + Simp[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] && Int 
egersQ[2*m, 2*p]
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71

\[-\frac {2 a^{5} \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \left (3+\sin \left (d x +c \right )\right )}{16 \left (-a +a \sin \left (d x +c \right )\right )^{2}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 a^{\frac {3}{2}}}\right )}{d}\]

Input:

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x)
 

Output:

-2*a^5*(-1/16*(a+a*sin(d*x+c))^(1/2)*(3+sin(d*x+c))/(-a+a*sin(d*x+c))^2+1/ 
32/a^(3/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.31 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) - 4 \, \sqrt {\frac {1}{2}} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 2 \, {\left (a^{3} \sin \left (d x + c\right ) + 3 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \] Input:

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

1/16*(sqrt(1/2)*(a^3*cos(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^3)*sqrt(a)* 
log(-(a*sin(d*x + c) - 4*sqrt(1/2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a) 
/(sin(d*x + c) - 1)) - 2*(a^3*sin(d*x + c) + 3*a^3)*sqrt(a*sin(d*x + c) + 
a))/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.25 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {\sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left ({\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 2 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{6}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}}{32 \, a d} \] Input:

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

1/32*(sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(s 
qrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a))) + 4*((a*sin(d*x + c) + a)^(3/2 
)*a^5 + 2*sqrt(a*sin(d*x + c) + a)*a^6)/((a*sin(d*x + c) + a)^2 - 4*(a*sin 
(d*x + c) + a)*a + 4*a^2))/(a*d)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.41 \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {\sqrt {2} a^{\frac {7}{2}} {\left (\frac {4 \, {\left (\frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}^{2} - 4} - \log \left ({\left | \frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) + \log \left ({\left | \frac {1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{64 \, d} \] Input:

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

1/64*sqrt(2)*a^(7/2)*(4*(1/cos(-1/4*pi + 1/2*d*x + 1/2*c) + cos(-1/4*pi + 
1/2*d*x + 1/2*c))/((1/cos(-1/4*pi + 1/2*d*x + 1/2*c) + cos(-1/4*pi + 1/2*d 
*x + 1/2*c))^2 - 4) - log(abs(1/cos(-1/4*pi + 1/2*d*x + 1/2*c) + cos(-1/4* 
pi + 1/2*d*x + 1/2*c) + 2)) + log(abs(1/cos(-1/4*pi + 1/2*d*x + 1/2*c) + c 
os(-1/4*pi + 1/2*d*x + 1/2*c) - 2)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^5} \,d x \] Input:

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^5,x)
 

Output:

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^5, x)
 

Reduce [F]

\[ \int \sec ^5(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\sqrt {a}\, a^{3} \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )^{3}d x +3 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )^{2}d x \right )+3 \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5} \sin \left (d x +c \right )d x \right )+\int \sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}d x \right ) \] Input:

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^(7/2),x)
 

Output:

sqrt(a)*a**3*(int(sqrt(sin(c + d*x) + 1)*sec(c + d*x)**5*sin(c + d*x)**3,x 
) + 3*int(sqrt(sin(c + d*x) + 1)*sec(c + d*x)**5*sin(c + d*x)**2,x) + 3*in 
t(sqrt(sin(c + d*x) + 1)*sec(c + d*x)**5*sin(c + d*x),x) + int(sqrt(sin(c 
+ d*x) + 1)*sec(c + d*x)**5,x))