Integrand size = 33, antiderivative size = 136 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=-\frac {2 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 \left (a^2 A+3 A b^2-3 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^3 d}-\frac {2 b^2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a+b) d}+\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d} \]
-2*(A*b-B*a)*(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^(1/2))/a^2/d+2/3*(A*a^2+3*A*b^2-3*B*a*b)*(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^(1/2))/a^3 /d-2*b^2*(A*b-B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^3/(a+b)/d+2/3*A*sin(d*x+c)*cos (d*x+c)^(1/2)/a/d
Time = 1.79 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\frac {\frac {(-A b+3 a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+A \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+2 A \sqrt {\cos (c+d x)} \sin (c+d x)+\frac {3 (-A b+a B) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{3 a d} \]
(((-(A*b) + 3*a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + A* (2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)) + 2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x] + (3*(-(A*b) + a*B)*(- 2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[Ar cSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sq rt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/(3*a*d )
Time = 1.15 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3433, 3042, 3469, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 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.
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A \cos (c+d x)+B)}{a \cos (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{a \sin \left (c+d x+\frac {\pi }{2}\right )+b}dx\) |
| \(\Big \downarrow \) 3469 |
| \(\displaystyle \frac {2 \int \frac {-3 (A b-a B) \cos ^2(c+d x)+a A \cos (c+d x)+A b}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {-3 (A b-a B) \cos ^2(c+d x)+a A \cos (c+d x)+A b}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-3 (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a A \sin \left (c+d x+\frac {\pi }{2}\right )+A b}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \frac {-\frac {\int -\frac {a A b+\left (A a^2-3 b B a+3 A b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {3 (A b-a B) \int \sqrt {\cos (c+d x)}dx}{a}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {a A b+\left (A a^2-3 b B a+3 A b^2\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {3 (A b-a B) \int \sqrt {\cos (c+d x)}dx}{a}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a A b+\left (A a^2-3 b B a+3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {3 (A b-a B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {\int \frac {a A b+\left (A a^2-3 b B a+3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \frac {\frac {\frac {\left (a^2 A-3 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {3 b^2 (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (a^2 A-3 a b B+3 A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {3 b^2 (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2 A-3 a b B+3 A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {3 b^2 (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {\frac {\frac {2 \left (a^2 A-3 a b B+3 A b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {6 b^2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}}{a}-\frac {6 (A b-a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a d}\) |
((-6*(A*b - a*B)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((2*(a^2*A + 3*A*b^2 - 3*a*b*B)*EllipticF[(c + d*x)/2, 2])/(a*d) - (6*b^2*(A*b - a*B)*EllipticPi [(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a + b)*d))/a)/(3*a) + (2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*d)
3.6.75.3.1 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[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[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_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin [e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !(IGt Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (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 - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, 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. \(821\) vs. \(2(208)=416\).
Time = 9.56 (sec) , antiderivative size = 822, normalized size of antiderivative = 6.04
-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A*cos(1/2* d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^3-4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2 *c)^4*a^2*b-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3+2*A*cos(1/2*d* x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2*b+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2 ))*b^3+A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-A*a^2*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))+3 *A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF (cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-3*A*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*( 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))*a^2*b-3*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))*a*b^2-3*B*(si n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos( 1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*a*b^2-3*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/ 2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2* b+3*B*a*b^2*(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))-3*B*(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))*a^3...
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{b \sec \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]