\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [70]

Optimal result

Integrand size = 33, antiderivative size = 147 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {2 (7 A+9 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 b d \sqrt {\cos (c+d x)}}+\frac {2 A b^4 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {2 b^2 (7 A+9 C) \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {2 (7 A+9 C) \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}} \] Output:

-2/15*(7*A+9*C)*(b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)) 
/b/d/cos(d*x+c)^(1/2)+2/9*A*b^4*sin(d*x+c)/d/(b*cos(d*x+c))^(9/2)+2/45*b^2 
*(7*A+9*C)*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/15*(7*A+9*C)*sin(d*x+c)/d/( 
b*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.49 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {-6 (7 A+9 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 (7 A+9 C) \sin (c+d x)+2 \sec (c+d x) \left (7 A+9 C+5 A \sec ^2(c+d x)\right ) \tan (c+d x)}{45 d \sqrt {b \cos (c+d x)}} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5)/Sqrt[b*Cos[c + d*x]],x]
 

Output:

(-6*(7*A + 9*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 6*(7*A + 9* 
C)*Sin[c + d*x] + 2*Sec[c + d*x]*(7*A + 9*C + 5*A*Sec[c + d*x]^2)*Tan[c + 
d*x])/(45*d*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 2030, 3491, 3042, 3116, 3042, 3116, 3042, 3121, 3042, 3119}

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.

\(\displaystyle \int \frac {\sec ^5(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5 \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^5 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{11/2}}dx\)

\(\Big \downarrow \) 3491

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \int \frac {1}{(b \cos (c+d x))^{7/2}}dx}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \left (\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \left (\frac {3 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \cos (c+d x)}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b^5 \left (\frac {(7 A+9 C) \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 A \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^5)/Sqrt[b*Cos[c + d*x]],x]
 

Output:

b^5*((2*A*Sin[c + d*x])/(9*b*d*(b*Cos[c + d*x])^(9/2)) + ((7*A + 9*C)*((2* 
Sin[c + d*x])/(5*b*d*(b*Cos[c + d*x])^(5/2)) + (3*((-2*Sqrt[b*Cos[c + d*x] 
]*EllipticE[(c + d*x)/2, 2])/(b^2*d*Sqrt[Cos[c + d*x]]) + (2*Sin[c + d*x]) 
/(b*d*Sqrt[b*Cos[c + d*x]])))/(5*b^2)))/(9*b^2))
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3491

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x 
_)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m 
+ 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1))   Int[(b*Sin[e + f* 
x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(728\) vs. \(2(131)=262\).

Time = 2.70 (sec) , antiderivative size = 729, normalized size of antiderivative = 4.96

method	result	size
default	\(\text {Expression too large to display}\)	\(729\)
parts	\(\text {Expression too large to display}\)	\(779\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-(b*(-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*(-1/144*c 
os(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/ 
2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2* 
d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-14/ 
15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(b*(-1+2*cos(1/2*d*x+1/2*c)^2)* 
sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2* 
d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^( 
1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d 
*x+1/2*c)^2))^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1 
/2*d*x+1/2*c),2^(1/2))))+2/5*C/b/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c 
)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1 
/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4- 
24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^( 
1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*si 
n(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d* 
x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1 
/2*c),2^(1/2)))*(-2*b*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(1/2))/ 
sin(1/2*d*x+1/2*c)/(b*(-1+2*cos(1/2*d*x+1/2*c)^2))^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} {\left (7 i \, A + 9 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (-7 i \, A - 9 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, {\left (7 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (7 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{45 \, b d \cos \left (d x + c\right )^{5}} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x, algorith 
m="fricas")
 

Output:

-2/45*(3*sqrt(1/2)*(7*I*A + 9*I*C)*sqrt(b)*cos(d*x + c)^5*weierstrassZeta( 
-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt 
(1/2)*(-7*I*A - 9*I*C)*sqrt(b)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weier 
strassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (3*(7*A + 9*C)*cos 
(d*x + c)^4 + (7*A + 9*C)*cos(d*x + c)^2 + 5*A)*sqrt(b*cos(d*x + c))*sin(d 
*x + c))/(b*d*cos(d*x + c)^5)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**5/(b*cos(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{5}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x, algorith 
m="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^5/sqrt(b*cos(d*x + c)), x)
 

Giac [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{5}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x, algorith 
m="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^5/sqrt(b*cos(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^5\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^5*(b*cos(c + d*x))^(1/2)),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^5*(b*cos(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{5}d x \right ) c \right )}{b} \] Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^5/(b*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(b)*(int((sqrt(cos(c + d*x))*sec(c + d*x)**5)/cos(c + d*x),x)*a + int 
(sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**5,x)*c))/b