Integrand size = 23, antiderivative size = 95 \[ \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx=-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{d}+\frac {3 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {4 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {\sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{d} \]
3/7*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+ 1/2*c),2/7*14^(1/2))/d*7^(1/2)+4/7*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d* x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2,2/7*14^(1/2))/d*7^(1/2)-(cos(1/2* d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2/7*14 ^(1/2))/d*7^(1/2)+(3+4*cos(d*x+c))^(1/2)*tan(d*x+c)/d
Result contains complex when optimal does not.
Time = 10.87 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.65 \[ \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\frac {6 \sqrt {7} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )+\frac {i \sqrt {7} \left (21 E\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right )|-\frac {1}{7}\right )-12 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )-8 \operatorname {EllipticPi}\left (-\frac {1}{3},i \text {arcsinh}\left (\sqrt {3+4 \cos (c+d x)}\right ),-\frac {1}{7}\right )\right ) \sin (c+d x)}{\sqrt {\sin ^2(c+d x)}}+21 \sqrt {3+4 \cos (c+d x)} \tan (c+d x)}{21 d} \]
(6*Sqrt[7]*EllipticPi[2, (c + d*x)/2, 8/7] + (I*Sqrt[7]*(21*EllipticE[I*Ar cSinh[Sqrt[3 + 4*Cos[c + d*x]]], -1/7] - 12*EllipticF[I*ArcSinh[Sqrt[3 + 4 *Cos[c + d*x]]], -1/7] - 8*EllipticPi[-1/3, I*ArcSinh[Sqrt[3 + 4*Cos[c + d *x]]], -1/7])*Sin[c + d*x])/Sqrt[Sin[c + d*x]^2] + 21*Sqrt[3 + 4*Cos[c + d *x]]*Tan[c + d*x])/(21*d)
Time = 0.76 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3275, 27, 3042, 3539, 25, 3042, 3132, 3481, 3042, 3140, 3284}
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.
| \(\displaystyle \int \sqrt {4 \cos (c+d x)+3} \sec ^2(c+d x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3275 |
| \(\displaystyle \int \frac {2 \left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int \frac {\left (1-\cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int \frac {1-\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3539 |
| \(\displaystyle 2 \left (-\frac {1}{4} \int \sqrt {4 \cos (c+d x)+3}dx-\frac {1}{4} \int -\frac {(3 \cos (c+d x)+4) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 \left (\frac {1}{4} \int \frac {(3 \cos (c+d x)+4) \sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx-\frac {1}{4} \int \sqrt {4 \cos (c+d x)+3}dx\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {1}{4} \int \frac {3 \sin \left (c+d x+\frac {\pi }{2}\right )+4}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx-\frac {1}{4} \int \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}dx\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle 2 \left (\frac {1}{4} \int \frac {3 \sin \left (c+d x+\frac {\pi }{2}\right )+4}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{2 d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle 2 \left (\frac {1}{4} \left (3 \int \frac {1}{\sqrt {4 \cos (c+d x)+3}}dx+4 \int \frac {\sec (c+d x)}{\sqrt {4 \cos (c+d x)+3}}dx\right )-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{2 d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {1}{4} \left (3 \int \frac {1}{\sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx+4 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx\right )-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{2 d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle 2 \left (\frac {1}{4} \left (4 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {4 \sin \left (c+d x+\frac {\pi }{2}\right )+3}}dx+\frac {6 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}\right )-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{2 d}\right )+\frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {\sqrt {4 \cos (c+d x)+3} \tan (c+d x)}{d}+2 \left (\frac {1}{4} \left (\frac {6 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}+\frac {8 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {8}{7}\right )}{\sqrt {7} d}\right )-\frac {\sqrt {7} E\left (\frac {1}{2} (c+d x)|\frac {8}{7}\right )}{2 d}\right )\) |
2*(-1/2*(Sqrt[7]*EllipticE[(c + d*x)/2, 8/7])/d + ((6*EllipticF[(c + d*x)/ 2, 8/7])/(Sqrt[7]*d) + (8*EllipticPi[2, (c + d*x)/2, 8/7])/(Sqrt[7]*d))/4) + (Sqrt[3 + 4*Cos[c + d*x]]*Tan[c + d*x])/d
3.6.14.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^(m + 1)*((c + d*Sin[e + f*x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^ (n - 1)*Simp[a*c*(m + 1) + b*d*n + (a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(m + n + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && LtQ[0, n, 1] && IntegersQ[2*m, 2*n]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp [C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x], x] - Simp[1/(b*d) Int[Simp[a *c*C - A*b*d + (b*c*C + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*( c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b *c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(165)=330\).
Time = 3.78 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.68
| method | result | size |
| default | \(-\frac {\sqrt {-\left (1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}+\frac {3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2 \sqrt {2}\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}-\frac {4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, 2 \sqrt {2}\right )}{\sqrt {-8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(350\) |
-(-(1-8*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+ 1/2*c)*(-8*sin(1/2*d*x+1/2*c)^4+7*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d *x+1/2*c)^2-1)+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-8*cos(1/2*d*x+1/2*c)^2)^( 1/2)/(-8*sin(1/2*d*x+1/2*c)^4+7*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos( 1/2*d*x+1/2*c),2*2^(1/2))+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-8*cos(1/2*d*x+1/ 2*c)^2)^(1/2)/(-8*sin(1/2*d*x+1/2*c)^4+7*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip ticE(cos(1/2*d*x+1/2*c),2*2^(1/2))-4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-8*cos (1/2*d*x+1/2*c)^2)^(1/2)/(-8*sin(1/2*d*x+1/2*c)^4+7*sin(1/2*d*x+1/2*c)^2)^ (1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,2*2^(1/2)))/sin(1/2*d*x+1/2*c)/(8*co s(1/2*d*x+1/2*c)^2-1)^(1/2)/d
\[ \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int { \sqrt {4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2} \,d x } \]
\[ \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int \sqrt {4 \cos {\left (c + d x \right )} + 3} \sec ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int { \sqrt {4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2} \,d x } \]
\[ \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int { \sqrt {4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \sqrt {3+4 \cos (c+d x)} \sec ^2(c+d x) \, dx=\int \frac {\sqrt {4\,\cos \left (c+d\,x\right )+3}}{{\cos \left (c+d\,x\right )}^2} \,d x \]