Integrand size = 33, antiderivative size = 209 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a^{3/2} (75 A+88 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+88 B) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (75 A+88 B) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (9 A+8 B) \cos ^2(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a A \cos ^3(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d} \]
1/64*a^(3/2)*(75*A+88*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2)) /d+1/64*a^2*(75*A+88*B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/96*a^2*(75*A +88*B)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/24*a^2*(9*A+8*B)*c os(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a*A*cos(d*x+c)^3*sin(d *x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 0.91 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\frac {a \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left ((285 A+296 B+2 (93 A+88 B) \cos (c+d x)+4 (15 A+8 B) \cos (2 (c+d x))+12 A \cos (3 (c+d x))) \sqrt {1-\sec (c+d x)} \sin (c+d x)+3 (75 A+88 B) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)\right )}{192 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]
(a*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*((285*A + 296*B + 2*(93*A + 88* B)*Cos[c + d*x] + 4*(15*A + 8*B)*Cos[2*(c + d*x)] + 12*A*Cos[3*(c + d*x)]) *Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 3*(75*A + 88*B)*ArcTanh[Sqrt[1 - Se c[c + d*x]]]*Tan[c + d*x]))/(192*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x ]])
Time = 1.09 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4292, 3042, 4261, 216}
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.
\(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^{3/2} (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4505 |
\(\displaystyle \frac {1}{4} \int \frac {1}{2} \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} (a (9 A+8 B)+a (5 A+8 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int \cos ^3(c+d x) \sqrt {\sec (c+d x) a+a} (a (9 A+8 B)+a (5 A+8 B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (9 A+8 B)+a (5 A+8 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 4503 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 A+88 B) \int \cos ^2(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 A+88 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 A+88 B) \left (\frac {3}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 A+88 B) \left (\frac {3}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 4292 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 A+88 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 A+88 B) \left (\frac {3}{4} \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 4261 |
\(\displaystyle \frac {1}{8} \left (\frac {1}{6} a (75 A+88 B) \left (\frac {3}{4} \left (\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{8} \left (\frac {a^2 (9 A+8 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {1}{6} a (75 A+88 B) \left (\frac {3}{4} \left (\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a A \sin (c+d x) \cos ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{4 d}\) |
(a*A*Cos[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(4*d) + ((a^2*( 9*A + 8*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + ( a*(75*A + 88*B)*((a*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x ]]) + (3*((Sqrt[a]*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]] )/d + (a*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4))/6)/8
3.2.34.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a *B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
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[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(185)=370\).
Time = 30.54 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.92
method | result | size |
default | \(\frac {a \left (48 A \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+120 A \sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}+64 B \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )+225 A \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+150 A \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+264 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+176 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+225 A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+225 A \cos \left (d x +c \right ) \sin \left (d x +c \right )+264 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+264 B \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{192 d \left (\cos \left (d x +c \right )+1\right )}\) | \(401\) |
1/192*a/d*(48*A*cos(d*x+c)^4*sin(d*x+c)+120*A*sin(d*x+c)*cos(d*x+c)^3+64*B *cos(d*x+c)^3*sin(d*x+c)+225*A*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x +c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+1 50*A*cos(d*x+c)^2*sin(d*x+c)+264*B*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos (d*x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+ c)+176*B*sin(d*x+c)*cos(d*x+c)^2+225*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+225* A*cos(d*x+c)*sin(d*x+c)+264*B*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+ c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+264*B*cos(d*x +c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 0.37 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.89 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 88 \, B\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (48 \, A a \cos \left (d x + c\right )^{4} + 8 \, {\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 88 \, B\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (48 \, A a \cos \left (d x + c\right )^{4} + 8 \, {\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A + 88 \, B\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
[1/384*(3*((75*A + 88*B)*a*cos(d*x + c) + (75*A + 88*B)*a)*sqrt(-a)*log((2 *a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos (d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(48*A *a*cos(d*x + c)^4 + 8*(15*A + 8*B)*a*cos(d*x + c)^3 + 2*(75*A + 88*B)*a*co s(d*x + c)^2 + 3*(75*A + 88*B)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/192*(3*((75*A + 88*B)* a*cos(d*x + c) + (75*A + 88*B)*a)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a) /cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a*cos(d*x + c) ^4 + 8*(15*A + 8*B)*a*cos(d*x + c)^3 + 2*(75*A + 88*B)*a*cos(d*x + c)^2 + 3*(75*A + 88*B)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si n(d*x + c))/(d*cos(d*x + c) + d)]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
\[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^4\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]