Integrand size = 42, antiderivative size = 207 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(5 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 B-43 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(B-C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 B-7 C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(35 B-11 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:
-(5*B-2*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+1/3 2*(115*B-43*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2 ))*2^(1/2)/a^(5/2)/d-1/4*(B-C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(1 5*B-7*C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/16*(35*B-11*C)*sin(d*x+c) /a^2/d/(a+a*sec(d*x+c))^(1/2)
Time = 3.22 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {-128 (5 B-2 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+4 \sqrt {2} (115 B-43 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+\sqrt {1-\sec (c+d x)} (55 B-15 C+(43 B-11 C+8 B \cos (2 (c+d x))) \sec (c+d x)) \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \] Input:
Integrate[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[ c + d*x])^(5/2),x]
Output:
(-128*(5*B - 2*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^5*Sec[c + d*x]^3*Sin[(c + d*x)/2] + 4*Sqrt[2]*(115*B - 43*C)*ArcTanh[Sqrt[1 - Sec [c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^5*Sec[c + d*x]^3*Sin[(c + d*x)/2] + S qrt[1 - Sec[c + d*x]]*(55*B - 15*C + (43*B - 11*C + 8*B*Cos[2*(c + d*x)])* Sec[c + d*x])*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d *x]))^(5/2))
Time = 1.52 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.08, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4560, 3042, 4508, 27, 3042, 4508, 27, 3042, 4510, 25, 3042, 4408, 3042, 4261, 216, 4282, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\cos (c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (2 a (5 B-C)-5 a (B-C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (2 a (5 B-C)-5 a (B-C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a (5 B-C)-5 a (B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 a^2 (35 B-11 C)-3 a^2 (15 B-7 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 a^2 (35 B-11 C)-3 a^2 (15 B-7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 a^2 (35 B-11 C)-3 a^2 (15 B-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {16 a^3 (5 B-2 C)-a^3 (35 B-11 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {16 a^3 (5 B-2 C)-a^3 (35 B-11 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {16 a^3 (5 B-2 C)-a^3 (35 B-11 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {16 a^2 (5 B-2 C) \int \sqrt {\sec (c+d x) a+a}dx-a^3 (115 B-43 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {16 a^2 (5 B-2 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^3 (115 B-43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^3 (115 B-43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {32 a^3 (5 B-2 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {32 a^{5/2} (5 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^3 (115 B-43 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (115 B-43 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {32 a^{5/2} (5 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (35 B-11 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {32 a^{5/2} (5 B-2 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{5/2} (115 B-43 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 B-7 C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
Input:
Int[(Cos[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d* x])^(5/2),x]
Output:
-1/4*((B - C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a*(15* B - 7*C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (-(((32*a^(5/2)*(5 *B - 2*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sq rt[2]*a^(5/2)*(115*B - 43*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a) + (2*a^2*(35*B - 11*C)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a^2))/(8*a^2)
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_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c/a Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[(b*c - a*d)/a Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 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*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d *n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
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. \(503\) vs. \(2(178)=356\).
Time = 4.07 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.43
| method | result | size |
| default | \(-\frac {\left (\left (80 \cos \left (d x +c \right )^{3}+240 \cos \left (d x +c \right )^{2}+240 \cos \left (d x +c \right )+80\right ) \sqrt {2}\, B \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-32 \cos \left (d x +c \right )^{3}-96 \cos \left (d x +c \right )^{2}-96 \cos \left (d x +c \right )-32\right ) \sqrt {2}\, C \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-115 \cos \left (d x +c \right )^{3}-345 \cos \left (d x +c \right )^{2}-345 \cos \left (d x +c \right )-115\right ) B \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (43 \cos \left (d x +c \right )^{3}+129 \cos \left (d x +c \right )^{2}+129 \cos \left (d x +c \right )+43\right ) C \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-32 \cos \left (d x +c \right )^{2}-110 \cos \left (d x +c \right )-70\right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (30 \cos \left (d x +c \right )+22\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right )}\) | \(504\) |
Input:
int(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,me thod=_RETURNVERBOSE)
Output:
-1/32/d/a^3*((80*cos(d*x+c)^3+240*cos(d*x+c)^2+240*cos(d*x+c)+80)*2^(1/2)* B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)/(csc(d*x+c)^2-2*csc (d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(-32*cos (d*x+c)^3-96*cos(d*x+c)^2-96*cos(d*x+c)-32)*2^(1/2)*C*(-2*cos(d*x+c)/(cos( d*x+c)+1))^(1/2)*arctanh(2^(1/2)/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot (d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(-115*cos(d*x+c)^3-345*cos(d*x +c)^2-345*cos(d*x+c)-115)*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*co s(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(43*cos(d*x+c)^3+129 *cos(d*x+c)^2+129*cos(d*x+c)+43)*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln ((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+sin(d*x+c)*co s(d*x+c)*(-32*cos(d*x+c)^2-110*cos(d*x+c)-70)*B+sin(d*x+c)*cos(d*x+c)*(30* cos(d*x+c)+22)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3* cos(d*x+c)+1)
Time = 4.22 (sec) , antiderivative size = 739, normalized size of antiderivative = 3.57 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="fricas")
Output:
[1/64*(sqrt(2)*((115*B - 43*C)*cos(d*x + c)^3 + 3*(115*B - 43*C)*cos(d*x + c)^2 + 3*(115*B - 43*C)*cos(d*x + c) + 115*B - 43*C)*sqrt(-a)*log(-(2*sqr t(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos (d*x + c) + 1)) + 32*((5*B - 2*C)*cos(d*x + c)^3 + 3*(5*B - 2*C)*cos(d*x + c)^2 + 3*(5*B - 2*C)*cos(d*x + c) + 5*B - 2*C)*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)*s in(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 4*(16*B*cos(d*x + c)^3 + 5*(11*B - 3*C)*cos(d*x + c)^2 + (35*B - 11*C)*cos(d*x + c))*sqrt((a *cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a ^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((115* B - 43*C)*cos(d*x + c)^3 + 3*(115*B - 43*C)*cos(d*x + c)^2 + 3*(115*B - 43 *C)*cos(d*x + c) + 115*B - 43*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 32*((5*B - 2* C)*cos(d*x + c)^3 + 3*(5*B - 2*C)*cos(d*x + c)^2 + 3*(5*B - 2*C)*cos(d*x + c) + 5*B - 2*C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*co s(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*(16*B*cos(d*x + c)^3 + 5*(11*B - 3* C)*cos(d*x + c)^2 + (35*B - 11*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/ cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(cos(d*x+c)**2*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**( 5/2),x)
Output:
Integral((B + C*sec(c + d*x))*cos(c + d*x)**2*sec(c + d*x)/(a*(sec(c + d*x ) + 1))**(5/2), x)
\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^2/(a*sec(d*x + c) + a)^(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (178) = 356\).
Time = 1.07 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.24 \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2 ),x, algorithm="giac")
Output:
-1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(B*a^5 - C*a^5)*ta n(1/2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x + c))) - sqrt(2)*(21*B*a^5 - 13*C*a^ 5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt(2)*(115*B - 43*C)* log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^ 2)/(sqrt(-a)*a^2*sgn(cos(d*x + c))) - 32*(5*B - 2*C)*log(abs(-562949953421 312*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^ 2 - 1125899906842624*sqrt(2)*abs(a) + 1688849860263936*a)/abs(-56294995342 1312*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) ^2 + 1125899906842624*sqrt(2)*abs(a) + 1688849860263936*a))/(sqrt(-a)*a*ab s(a)*sgn(cos(d*x + c))) - 128*sqrt(2)*(3*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B - B*a)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)*sqrt(-a)*a*sgn( cos(d*x + c))))/d
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((cos(c + d*x)^2*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d* x))^(5/2),x)
Output:
int((cos(c + d*x)^2*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d* x))^(5/2), x)
\[ \int \frac {\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b \right )}{a^{3}} \] Input:
int(cos(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**2)/(se c(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c + int((sqrt(s ec(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x))/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*b))/a**3