\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [270]

Optimal result

Integrand size = 41, antiderivative size = 209 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {6 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b d \sqrt {\cos (c+d x)}}+\frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^3 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}}+\frac {2 b^2 B \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b (5 A+7 C) \sin (c+d x)}{21 d (b \cos (c+d x))^{3/2}}+\frac {6 B \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \] Output:

-6/5*B*(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)+2/21*(5*A+7*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c 
,2^(1/2))/d/(b*cos(d*x+c))^(1/2)+2/7*A*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(7/ 
2)+2/5*b^2*B*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/21*b*(5*A+7*C)*sin(d*x+c) 
/d/(b*cos(d*x+c))^(3/2)+6/5*B*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.64 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \left (-63 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+63 B \sin (c+d x)+25 A \tan (c+d x)+35 C \tan (c+d x)+21 B \sec (c+d x) \tan (c+d x)+15 A \sec ^2(c+d x) \tan (c+d x)\right )}{105 d \sqrt {b \cos (c+d x)}} \] Input:

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[b* 
Cos[c + d*x]],x]
 

Output:

(2*(-63*B*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*(5*A + 7*C)*Sqr 
t[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 63*B*Sin[c + d*x] + 25*A*Tan[c 
 + d*x] + 35*C*Tan[c + d*x] + 21*B*Sec[c + d*x]*Tan[c + d*x] + 15*A*Sec[c 
+ d*x]^2*Tan[c + d*x]))/(105*d*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.11, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 2030, 3500, 27, 3042, 3227, 3042, 3116, 3042, 3116, 3042, 3121, 3042, 3119, 3120}

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.

\(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^4 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{9/2}}dx\)

\(\Big \downarrow \) 3500

\(\displaystyle b^4 \left (\frac {2 \int \frac {7 B b^2+(5 A+7 C) \cos (c+d x) b^2}{2 (b \cos (c+d x))^{7/2}}dx}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle b^4 \left (\frac {\int \frac {7 B b^2+(5 A+7 C) \cos (c+d x) b^2}{(b \cos (c+d x))^{7/2}}dx}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {\int \frac {7 B b^2+(5 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \int \frac {1}{(b \cos (c+d x))^{5/2}}dx+7 b^2 B \int \frac {1}{(b \cos (c+d x))^{7/2}}dx}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx+7 b^2 B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {\int \frac {1}{\sqrt {b \cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \cos (c+d x)}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b^4 \left (\frac {b (5 A+7 C) \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )+7 b^2 B \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{7 b^3}+\frac {2 A \sin (c+d x)}{7 b d (b \cos (c+d x))^{7/2}}\right )\)

Input:

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[b*Cos[c 
+ d*x]],x]
 

Output:

b^4*((2*A*Sin[c + d*x])/(7*b*d*(b*Cos[c + d*x])^(7/2)) + (b*(5*A + 7*C)*(( 
2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*b^2*d*Sqrt[b*Cos[c + d* 
x]]) + (2*Sin[c + d*x])/(3*b*d*(b*Cos[c + d*x])^(3/2))) + 7*b^2*B*((2*Sin[ 
c + d*x])/(5*b*d*(b*Cos[c + d*x])^(5/2)) + (3*((-2*Sqrt[b*Cos[c + d*x]]*El 
lipticE[(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)))/(7*b^3))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3500

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(725\) vs. \(2(184)=368\).

Time = 1.81 (sec) , antiderivative size = 726, normalized size of antiderivative = 3.47

method	result	size
default	\(\text {Expression too large to display}\)	\(726\)
parts	\(\text {Expression too large to display}\)	\(1001\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(b*cos(d*x+c))^(1/2),x,me 
thod=_RETURNVERBOSE)
 

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*(-1/56*co 
s(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)^4-5/42*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)^2+5/21* 
(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)))+2/5*B/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))*sin(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)+2*C*(-1/6*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)^2+1/3*(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))))/sin(1/2*d*x+1/2*c)/(b*(-1+2*co 
s(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 = 234, normalized size of antiderivative = 1.12 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {2 \, {\left (5 \, \sqrt {\frac {1}{2}} {\left (5 i \, A + 7 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {\frac {1}{2}} {\left (-5 i \, A - 7 i \, C\right )} \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 i \, \sqrt {\frac {1}{2}} B \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 i \, \sqrt {\frac {1}{2}} B \sqrt {b} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (63 \, B \cos \left (d x + c\right )^{3} + 5 \, {\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 21 \, B \cos \left (d x + c\right ) + 15 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, b d \cos \left (d x + c\right )^{4}} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(b*cos(d*x+c))^(1/2 
),x, algorithm="fricas")
 

Output:

-2/105*(5*sqrt(1/2)*(5*I*A + 7*I*C)*sqrt(b)*cos(d*x + c)^4*weierstrassPInv 
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(1/2)*(-5*I*A - 7*I*C)* 
sqrt(b)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
 + c)) + 63*I*sqrt(1/2)*B*sqrt(b)*cos(d*x + c)^4*weierstrassZeta(-4, 0, we 
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 63*I*sqrt(1/2)* 
B*sqrt(b)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
 cos(d*x + c) - I*sin(d*x + c))) - (63*B*cos(d*x + c)^3 + 5*(5*A + 7*C)*co 
s(d*x + c)^2 + 21*B*cos(d*x + c) + 15*A)*sqrt(b*cos(d*x + c))*sin(d*x + c) 
)/(b*d*cos(d*x + c)^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(b*cos(d*x+c))**( 
1/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(b*cos(d*x+c))^(1/2 
),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^4/sqrt(b*co 
s(d*x + c)), x)
 

Giac [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{4}}{\sqrt {b \cos \left (d x + c\right )}} \,d x } \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(b*cos(d*x+c))^(1/2 
),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^4/sqrt(b*co 
s(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^4\,\sqrt {b\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(b*cos(c + d*x 
))^(1/2)),x)
 

Output:

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(b*cos(c + d*x 
))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {b}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}}{\cos \left (d x +c \right )}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) b \right )}{b} \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(b*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(b)*(int((sqrt(cos(c + d*x))*sec(c + d*x)**4)/cos(c + d*x),x)*a + int 
(sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**4,x)*c + int(sqrt(cos(c + d 
*x))*sec(c + d*x)**4,x)*b))/b