Integrand size = 42, antiderivative size = 119 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3/2} (7 B+12 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^2 (5 B+4 C) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a B \cos (c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d} \]
1/4*a^(3/2)*(7*B+12*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d +1/4*a^2*(5*B+4*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*a*B*cos(d*x+c)* sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d
Time = 1.62 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \cos (c+d x) \sqrt {a (1+\sec (c+d x))} \left ((7 B+4 C+2 B \cos (c+d x)) \sqrt {1-\sec (c+d x)} \sin (c+d x)+(7 B+12 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)\right )}{4 d (1+\cos (c+d x)) \sqrt {1-\sec (c+d x)}} \]
(a*Cos[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*((7*B + 4*C + 2*B*Cos[c + d*x]) *Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + (7*B + 12*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x]))/(4*d*(1 + Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])
Time = 0.79 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4560, 3042, 4505, 27, 3042, 4503, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^{3/2} (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{2} \cos (c+d x) \sqrt {\sec (c+d x) a+a} (a (5 B+4 C)+a (B+4 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \cos (c+d x) \sqrt {\sec (c+d x) a+a} (a (5 B+4 C)+a (B+4 C) \sec (c+d x))dx+\frac {a B \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (5 B+4 C)+a (B+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a B \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a (7 B+12 C) \int \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (5 B+4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} a (7 B+12 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (5 B+4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {1}{4} \left (\frac {a^2 (5 B+4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (7 B+12 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a B \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{4} \left (\frac {a^{3/2} (7 B+12 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^2 (5 B+4 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \cos (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\) |
(a*B*Cos[c + d*x]*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) + ((a^(3/2) *(7*B + 12*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^2*(5*B + 4*C)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/4
3.4.71.3.1 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_)]*(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]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(332\) vs. \(2(103)=206\).
Time = 1.70 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.80
| method | result | size |
| default | \(\frac {a \left (7 B \,\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2 B \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}+12 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+7 B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+7 B \cos \left (d x +c \right ) \sin \left (d x +c \right )+12 C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+4 C \cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \left (\cos \left (d x +c \right )+1\right )}\) | \(333\) |
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,me thod=_RETURNVERBOSE)
1/4*a/d*(7*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)+2*B*sin(d*x+c)*cos( d*x+c)^2+12*C*(-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))*cos(d*x+c)+7*B*(-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))+7*B*cos(d*x+c)*sin(d*x+c)+12*C*(-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))+4*C*cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 0.32 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.69 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {{\left ({\left (7 \, B + 12 \, C\right )} a \cos \left (d x + c\right ) + {\left (7 \, B + 12 \, C\right )} a\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, B a \cos \left (d x + c\right )^{2} + {\left (7 \, B + 4 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {{\left ({\left (7 \, B + 12 \, C\right )} a \cos \left (d x + c\right ) + {\left (7 \, B + 12 \, C\right )} a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, B a \cos \left (d x + c\right )^{2} + {\left (7 \, B + 4 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
[1/8*(((7*B + 12*C)*a*cos(d*x + c) + (7*B + 12*C)*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*(2*B*a*cos( d*x + c)^2 + (7*B + 4*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d), -1/4*(((7*B + 12*C)*a*cos(d*x + c) + (7*B + 12*C)*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))) - (2*B*a*cos(d*x + c)^2 + (7*B + 4 *C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/ (d*cos(d*x + c) + d)]
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{3} \,d x } \]
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^3\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]