Integrand size = 31, antiderivative size = 169 \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=-\frac {6 A \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^2 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {6 A b \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \] Output:
-6/5*A*(b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d* x+c)^(1/2)+2/3*b*B*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)+2/5*A*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/3*b^ 2*B*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+6/5*A*b*sin(d*x+c)/d/(b*cos(d*x+c))^ (1/2)
Time = 0.69 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.63 \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {2 \sqrt {b \cos (c+d x)} \sec ^2(c+d x) \left (-9 A \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 B \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+5 B \sin (c+d x)+\frac {9}{2} A \sin (2 (c+d x))+3 A \tan (c+d x)\right )}{15 d} \] Input:
Integrate[Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
Output:
(2*Sqrt[b*Cos[c + d*x]]*Sec[c + d*x]^2*(-9*A*Cos[c + d*x]^(3/2)*EllipticE[ (c + d*x)/2, 2] + 5*B*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 5*B*S in[c + d*x] + (9*A*Sin[2*(c + d*x)])/2 + 3*A*Tan[c + d*x]))/(15*d)
Time = 0.76 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 2030, 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.
| \(\displaystyle \int \sec ^4(c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle b^4 \int \frac {A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle b^4 \left (A \int \frac {1}{(b \cos (c+d x))^{7/2}}dx+\frac {B \int \frac {1}{(b \cos (c+d x))^{5/2}}dx}{b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^4 \left (A \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx+\frac {B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{b}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle b^4 \left (A \left (\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^4 \left (A \left (\frac {3 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle b^4 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \cos (c+d x)}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^4 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle b^4 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^4 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle b^4 \left (\frac {B \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}+A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle b^4 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\) |
Input:
Int[Sqrt[b*Cos[c + d*x]]*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
Output:
b^4*((B*((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))))/b + A*( (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)))
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]
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]
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]
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]
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]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(575\) vs. \(2(148)=296\).
Time = 9.29 (sec) , antiderivative size = 576, normalized size of antiderivative = 3.41
| method | result | size |
| default | \(-\frac {2 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (72 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-36 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-20 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-72 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+36 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-20 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+20 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+10 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\) | \(576\) |
| parts | \(-\frac {2 A \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}}{5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}-\frac {2 B \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) b \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\) | \(605\) |
Input:
int((cos(d*x+c)*b)^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x,method=_RETURNVER BOSE)
Output:
-2/15*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2* d*x+1/2*c)^3/(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)*(72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-36*A*(sin(1/2*d *x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/ 2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/ 2*d*x+1/2*c)^4-72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+36*A*(sin(1/2* d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1 /2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+ 1/2*c)^4+20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^ (1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*A*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-9*A*(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))+10*B*cos(1/2 *d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1 /2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(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*cos(1/2*d*x+1/2*c)^2-1) *b)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.20 \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {\frac {1}{2}} B \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {\frac {1}{2}} B \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {\frac {1}{2}} A \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {\frac {1}{2}} A \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (9 \, A \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm= "fricas")
Output:
-2/15*(5*I*sqrt(1/2)*B*sqrt(b)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, c os(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(1/2)*B*sqrt(b)*cos(d*x + c)^3*wei erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*I*sqrt(1/2)*A*s qrt(b)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co s(d*x + c) + I*sin(d*x + c))) - 9*I*sqrt(1/2)*A*sqrt(b)*cos(d*x + c)^3*wei erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (9*A*cos(d*x + c)^2 + 5*B*cos(d*x + c) + 3*A)*sqrt(b*cos(d*x + c))* sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:
integrate((b*cos(d*x+c))**(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)
Output:
Timed out
\[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm= "maxima")
Output:
integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sec(d*x + c)^4, x)
\[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \sqrt {b \cos \left (d x + c\right )} \sec \left (d x + c\right )^{4} \,d x } \] Input:
integrate((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm= "giac")
Output:
integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sec(d*x + c)^4, x)
Timed out. \[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\int \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:
int(((b*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^4,x)
Output:
int(((b*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^4, x)
\[ \int \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\sqrt {b}\, \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{4}d x \right ) a \right ) \] Input:
int((b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)
Output:
sqrt(b)*(int(sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**4,x)*b + int(sq rt(cos(c + d*x))*sec(c + d*x)**4,x)*a)