\(\int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [111]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{8 \sqrt {2} d}-\frac {5 a^2 \cos (c+d x)}{8 d (a+a \sin (c+d x))^{3/2}}+\frac {5 a \sec (c+d x)}{6 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{3 d} \] Output:

-5/16*a^(1/2)*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2 
))*2^(1/2)/d-5/8*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)+5/6*a*sec(d*x+c)/ 
d/(a+a*sin(d*x+c))^(1/2)+1/3*sec(d*x+c)^3*(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 = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.39 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec ^3(c+d x) (a (1+\sin (c+d x)))^{3/2}}{6 a d} \] Input:

Integrate[Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(Hypergeometric2F1[-3/2, 2, -1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]^3*(a* 
(1 + Sin[c + d*x]))^(3/2))/(6*a*d)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3154, 3042, 3166, 3042, 3129, 3042, 3128, 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.

\(\displaystyle \int \sec ^4(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \sin (c+d x)+a}}{\cos (c+d x)^4}dx\)

\(\Big \downarrow \) 3154

\(\displaystyle \frac {5}{6} a \int \frac {\sec ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} a \int \frac {1}{\cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3166

\(\displaystyle \frac {5}{6} a \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} a \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {5}{6} a \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} a \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {5}{6} a \left (\frac {3}{2} a \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 a d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5}{6} a \left (\frac {3}{2} a \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )+\frac {\sec ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}\)

Input:

Int[Sec[c + d*x]^4*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(Sec[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(3*d) + (5*a*(Sec[c + d*x]/(d*Sq 
rt[a + a*Sin[c + d*x]]) + (3*a*(-1/2*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[ 
2]*Sqrt[a + a*Sin[c + d*x]])]/(Sqrt[2]*a^(3/2)*d) - Cos[c + d*x]/(2*d*(a + 
 a*Sin[c + d*x])^(3/2))))/2))/6
 

Defintions of rubi rules used

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 3128

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

rule 3129

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] + Simp[(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 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 = 157, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {15 \sqrt {2}\, \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sin \left (d x +c \right ) a -30 a^{\frac {5}{2}} \cos \left (d x +c \right )^{2}+15 \sqrt {2}\, \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -20 a^{\frac {5}{2}} \sin \left (d x +c \right )+4 a^{\frac {5}{2}}}{48 a^{\frac {3}{2}} \left (\sin \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(157\)

Input:

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/48/a^(3/2)*(15*2^(1/2)*(a-a*sin(d*x+c))^(3/2)*arctanh(1/2*(a-a*sin(d*x+c 
))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)*a-30*a^(5/2)*cos(d*x+c)^2+15*2^(1/2)* 
(a-a*sin(d*x+c))^(3/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)) 
*a-20*a^(5/2)*sin(d*x+c)+4*a^(5/2))/(sin(d*x+c)-1)/cos(d*x+c)/(a+a*sin(d*x 
+c))^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.33 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {15 \, \sqrt {\frac {1}{2}} \sqrt {a} \cos \left (d x + c\right )^{3} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {\frac {1}{2}} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{2} + 10 \, \sin \left (d x + c\right ) - 2\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{48 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

1/48*(15*sqrt(1/2)*sqrt(a)*cos(d*x + c)^3*log(-(a*cos(d*x + c)^2 - 4*sqrt( 
1/2)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 
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)) + 2*(15*cos(d*x 
 + c)^2 + 10*sin(d*x + c) - 2)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^3 
)
 

Sympy [F]

\[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \sec ^{4}{\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(a*(sin(c + d*x) + 1))*sec(c + d*x)**4, x)
 

Maxima [F]

\[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(a*sin(d*x + c) + a)*sec(d*x + c)^4, x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.30 \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (15 \, \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 15 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - \frac {4 \, {\left (6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}\right )} \sqrt {a}}{96 \, d} \] Input:

integrate(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

1/96*sqrt(2)*(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)) - 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)) - 6*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) - 4*(6*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)^3)*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^4,x)
 

Output:

int((a + a*sin(c + d*x))^(1/2)/cos(c + d*x)^4, x)
 

Reduce [F]

\[ \int \sec ^4(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) \] Input:

int(sec(d*x+c)^4*(a+a*sin(d*x+c))^(1/2),x)
 

Output:

sqrt(a)*int(sqrt(sin(c + d*x) + 1)*sec(c + d*x)**4,x)