\(\int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [112]

Optimal result

Integrand size = 23, antiderivative size = 149 \[ \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {35 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}-\frac {35 a^2}{96 d (a+a \sin (c+d x))^{3/2}}-\frac {35 a}{64 d \sqrt {a+a \sin (c+d x)}}+\frac {7 a \sec ^2(c+d x)}{16 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^4(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d} \] Output:

35/128*2^(1/2)*a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)) 
/d-35/96*a^2/d/(a+a*sin(d*x+c))^(3/2)-35/64*a/d/(a+a*sin(d*x+c))^(1/2)+7/1 
6*a*sec(d*x+c)^2/d/(a+a*sin(d*x+c))^(1/2)+1/4*sec(d*x+c)^4*(a+a*sin(d*x+c) 
)^(1/2)/d
 

Mathematica [C] (verified)

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

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.30 \[ \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {a^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},3,-\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{12 d (a+a \sin (c+d x))^{3/2}} \] Input:

Integrate[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

-1/12*(a^2*Hypergeometric2F1[-3/2, 3, -1/2, (1 + Sin[c + d*x])/2])/(d*(a + 
 a*Sin[c + d*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3154, 3042, 3166, 3042, 3146, 61, 61, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3154

\(\displaystyle \frac {7}{8} a \int \frac {\sec ^3(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{8} a \int \frac {1}{\cos (c+d x)^3 \sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3166

\(\displaystyle \frac {7}{8} a \left (\frac {5}{4} a \int \frac {\sec (c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{8} a \left (\frac {5}{4} a \int \frac {1}{\cos (c+d x) (\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 3146

\(\displaystyle \frac {7}{8} a \left (\frac {5 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{5/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {7}{8} a \left (\frac {5 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {7}{8} a \left (\frac {5 a^2 \left (\frac {\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {7}{8} a \left (\frac {5 a^2 \left (\frac {\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {7}{8} a \left (\frac {5 a^2 \left (\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {1}{3 a (a \sin (c+d x)+a)^{3/2}}\right )}{4 d}+\frac {\sec ^2(c+d x)}{2 d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^4(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d}\)

Input:

Int[Sec[c + d*x]^5*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]])/(4*d) + (7*a*(Sec[c + d*x]^2/(2* 
d*Sqrt[a + a*Sin[c + d*x]]) + (5*a^2*(-1/3*1/(a*(a + a*Sin[c + d*x])^(3/2) 
) + (ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]]/(Sqrt[2]*a^(3/2)) - 1/(a*Sqrt 
[a + a*Sin[c + d*x]]))/(2*a)))/(4*d)))/8
 

Defintions of rubi rules used

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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 3166

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)*S 
qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1)))   Int[(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]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {2 a^{5} \left (\frac {3}{16 a^{4} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {1}{24 a^{3} \left (a +a \sin \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a \left (11 \sin \left (d x +c \right )-15\right )}{8 \left (-a +a \sin \left (d x +c \right )\right )^{2}}-\frac {35 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}}{16 a^{4}}\right )}{d}\)	\(118\)

Input:

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2*a^5*(3/16/a^4/(a+a*sin(d*x+c))^(1/2)+1/24/a^3/(a+a*sin(d*x+c))^(3/2)+1/ 
16/a^4*(1/8*(a+a*sin(d*x+c))^(1/2)*a*(11*sin(d*x+c)-15)/(-a+a*sin(d*x+c))^ 
2-35/16*2^(1/2)/a^(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 = 121, normalized size of antiderivative = 0.81 \[ \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {105 \, \sqrt {\frac {1}{2}} \sqrt {a} \cos \left (d x + c\right )^{4} \log \left (-\frac {a \sin \left (d x + c\right ) + 4 \, \sqrt {\frac {1}{2}} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 2 \, {\left (35 \, \cos \left (d x + c\right )^{2} - 7 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 8\right )} \sin \left (d x + c\right ) + 8\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{384 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
 

Output:

1/384*(105*sqrt(1/2)*sqrt(a)*cos(d*x + c)^4*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*(35*c 
os(d*x + c)^2 - 7*(15*cos(d*x + c)^2 + 8)*sin(d*x + c) + 8)*sqrt(a*sin(d*x 
 + c) + a))/(d*cos(d*x + c)^4)
 

Sympy [F]

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

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(a*(sin(c + d*x) + 1))*sec(c + d*x)**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.13 \[ \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {105 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (105 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{2} - 350 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{3} + 224 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{4} + 64 \, a^{5}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 4 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2}}}{768 \, a d} \] Input:

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

-1/768*(105*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + 
a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a))) + 4*(105*(a*sin(d*x + c) 
 + a)^3*a^2 - 350*(a*sin(d*x + c) + a)^2*a^3 + 224*(a*sin(d*x + c) + a)*a^ 
4 + 64*a^5)/((a*sin(d*x + c) + a)^(7/2) - 4*(a*sin(d*x + c) + a)^(5/2)*a + 
 4*(a*sin(d*x + c) + a)^(3/2)*a^2))/(a*d)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.98 \[ \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {6 \, {\left (11 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 13 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {16 \, {\left (9 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} - 105 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 105 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{768 \, d} \] Input:

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

-1/768*sqrt(2)*sqrt(a)*(6*(11*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 13*cos(-1 
/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2 + 16*(9 
*cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 + 1)/cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 
105*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 105*log(-cos(-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. \[ \int \sec ^5(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^5} \,d x \] Input:

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^5,x)
 

Output:

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^5, x)
 

Reduce [F]

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

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2),x)
 

Output:

sqrt(a)*int(sqrt(sin(c + d*x) + 1)*sec(c + d*x)**5,x)