Integrand size = 33, antiderivative size = 237 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 \left (5 a^3 A-15 a A b^2-15 a^2 b B-3 b^3 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 b^2 (5 A b+9 a B) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}}+\frac {2 a^2 (5 a A-b B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 b B (b+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \] Output:
-2/5*(5*A*a^3-15*A*a*b^2-15*B*a^2*b-3*B*b^3)*cos(d*x+c)^(1/2)*EllipticE(si n(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/d+2/3*(9*A*a^2*b+A*b^3+3*B*a^3+ 3*B*a*b^2)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x +c)^(1/2)/d+2/15*b^2*(5*A*b+9*B*a)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*a^2*( 5*A*a-B*b)*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2/5*b*B*(b+a*sec(d*x+c))^2*sin(d* x+c)/d/sec(d*x+c)^(3/2)
Time = 10.85 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.73 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (12 \left (-5 a^3 A+15 a A b^2+15 a^2 b B+3 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 \left (9 a^2 A b+A b^3+3 a^3 B+3 a b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {2 \left (10 b^2 (A b+3 a B) \cos (c+d x)+3 \left (10 a^3 A+b^3 B+b^3 B \cos (2 (c+d x))\right )\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{30 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2),x ]
Output:
(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(12*(-5*a^3*A + 15*a*A*b^2 + 15*a^2 *b*B + 3*b^3*B)*EllipticE[(c + d*x)/2, 2] + 20*(9*a^2*A*b + A*b^3 + 3*a^3* B + 3*a*b^2*B)*EllipticF[(c + d*x)/2, 2] + (2*(10*b^2*(A*b + 3*a*B)*Cos[c + d*x] + 3*(10*a^3*A + b^3*B + b^3*B*Cos[2*(c + d*x)]))*Sin[c + d*x])/Sqrt [Cos[c + d*x]]))/(30*d)
Time = 1.69 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.02, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 3439, 3042, 4513, 27, 3042, 4562, 27, 3042, 4535, 3042, 4258, 3042, 3120, 4534, 3042, 4258, 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.
\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 3439 |
\(\displaystyle \int \frac {(a \sec (c+d x)+b)^3 (A \sec (c+d x)+B)}{\sec ^{\frac {5}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^3 \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4513 |
\(\displaystyle \frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2}{5} \int -\frac {(b+a \sec (c+d x)) \left (a (5 a A-b B) \sec ^2(c+d x)+\left (5 B a^2+10 A b a+3 b^2 B\right ) \sec (c+d x)+b (5 A b+9 a B)\right )}{2 \sec ^{\frac {3}{2}}(c+d x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {(b+a \sec (c+d x)) \left (a (5 a A-b B) \sec ^2(c+d x)+\left (5 B a^2+10 A b a+3 b^2 B\right ) \sec (c+d x)+b (5 A b+9 a B)\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \int \frac {\left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a (5 a A-b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 B a^2+10 A b a+3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+b (5 A b+9 a B)\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4562 |
\(\displaystyle \frac {1}{5} \left (\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2}{3} \int -\frac {3 a^2 (5 a A-b B) \sec ^2(c+d x)+5 \left (3 B a^3+9 A b a^2+3 b^2 B a+A b^3\right ) \sec (c+d x)+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{2 \sqrt {\sec (c+d x)}}dx\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 a^2 (5 a A-b B) \sec ^2(c+d x)+5 \left (3 B a^3+9 A b a^2+3 b^2 B a+A b^3\right ) \sec (c+d x)+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{\sqrt {\sec (c+d x)}}dx+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 a^2 (5 a A-b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (3 B a^3+9 A b a^2+3 b^2 B a+A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 (5 a A-b B) \sec ^2(c+d x)+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{\sqrt {\sec (c+d x)}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \int \sqrt {\sec (c+d x)}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 (5 a A-b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 (5 a A-b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 (5 a A-b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 a^2 (5 a A-b B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b \left (14 B a^2+15 A b a+3 b^2 B\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {6 a^2 (5 a A-b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^2 (5 a A-b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {6 a^2 (5 a A-b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-3 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {6 a^2 (5 a A-b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {6 a^2 (5 a A-b B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 \left (3 a^3 B+9 a^2 A b+3 a b^2 B+A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 \left (5 a^3 A-15 a^2 b B-15 a A b^2-3 b^3 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 b^2 (9 a B+5 A b) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 b B \sin (c+d x) (a \sec (c+d x)+b)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2),x]
Output:
(2*b*B*(b + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((2 *b^2*(5*A*b + 9*a*B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((-6*(5*a^3* A - 15*a*A*b^2 - 15*a^2*b*B - 3*b^3*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d *x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (10*(9*a^2*A*b + A*b^3 + 3*a^3*B + 3*a*b ^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (6*a^2*(5*a*A - b*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d)/3)/5
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[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* (x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + c*Csc[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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] + Sim p[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[ a*(a*B*n - A*b*(m - n - 1)) + (2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] & & LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(640\) vs. \(2(216)=432\).
Time = 13.83 (sec) , antiderivative size = 641, normalized size of antiderivative = 2.70
method | result | size |
default | \(-\frac {2 \left (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b^{3}+20 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b^{3}+60 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a \,b^{2}+24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b^{3}-30 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}-10 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}+45 A \,a^{2} 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 )+5 A \,b^{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 {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 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 ) a^{3}-45 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 ) a \,b^{2}-30 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{2}-6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}+15 a^{3} 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 )+15 B a \,b^{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 )-45 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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b -9 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 {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{3}\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(641\) |
parts | \(\text {Expression too large to display}\) | \(875\) |
Input:
int((a+cos(d*x+c)*b)^3*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x,method=_RETURNV ERBOSE)
Output:
-2/15*(-24*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b^3+20*A*cos(1/2*d*x+ 1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3+60*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c) ^4*a*b^2+24*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*b^3-30*A*cos(1/2*d*x +1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3-10*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c )^2*b^3+45*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))+5*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))+15*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)*El lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-45*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 ^2-30*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2-6*B*cos(1/2*d*x+1/2* c)*sin(1/2*d*x+1/2*c)^2*b^3+15*a^3*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))+15*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(co s(1/2*d*x+1/2*c),2^(1/2))-45*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-9*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)*EllipticE(cos(1/2*d *x+1/2*c),2^(1/2))*b^3)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2 )/d
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.14 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, B a^{3} + 9 i \, A a^{2} b + 3 i \, B a b^{2} + i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-3 i \, B a^{3} - 9 i \, A a^{2} b - 3 i \, B a b^{2} - i \, A b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (5 i \, A a^{3} - 15 i \, B a^{2} b - 15 i \, A a b^{2} - 3 i \, B b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-5 i \, A a^{3} + 15 i \, B a^{2} b + 15 i \, A a b^{2} + 3 i \, B b^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, B b^{3} \cos \left (d x + c\right )^{2} + 15 \, A a^{3} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorith m="fricas")
Output:
-1/15*(5*sqrt(2)*(3*I*B*a^3 + 9*I*A*a^2*b + 3*I*B*a*b^2 + I*A*b^3)*weierst rassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-3*I*B*a^3 - 9*I*A*a^2*b - 3*I*B*a*b^2 - I*A*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*sqrt(2)*(5*I*A*a^3 - 15*I*B*a^2*b - 15*I*A*a*b ^2 - 3*I*B*b^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-5*I*A*a^3 + 15*I*B*a^2*b + 15*I*A*a* b^2 + 3*I*B*b^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*B*b^3*cos(d*x + c)^2 + 15*A*a^3 + 5*(3*B*a *b^2 + A*b^3)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
Timed out. \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c)**(3/2),x)
Output:
Timed out
\[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorith m="maxima")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(3/2), x)
\[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorith m="giac")
Output:
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3*sec(d*x + c)^(3/2), x)
Timed out. \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int \left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \] Input:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^3,x)
Output:
int((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2)*(a + b*cos(c + d*x))^3, x)
\[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )d x \right ) a^{3} b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{4} \sec \left (d x +c \right )d x \right ) b^{4}+4 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )d x \right ) a \,b^{3}+6 \left (\int \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) a^{2} b^{2}+\left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) a^{4} \] Input:
int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x)
Output:
4*int(sqrt(sec(c + d*x))*cos(c + d*x)*sec(c + d*x),x)*a**3*b + int(sqrt(se c(c + d*x))*cos(c + d*x)**4*sec(c + d*x),x)*b**4 + 4*int(sqrt(sec(c + d*x) )*cos(c + d*x)**3*sec(c + d*x),x)*a*b**3 + 6*int(sqrt(sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x),x)*a**2*b**2 + int(sqrt(sec(c + d*x))*sec(c + d*x), x)*a**4