Integrand size = 43, antiderivative size = 229 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A-11 B+15 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(45 A-65 B+93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A-5 B+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-35 B+39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d} \] Output:
-1/4*(7*A-11*B+15*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^(3/2)/d-1/2*(A-B+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec( d*x+c))^(3/2)+1/15*(45*A-65*B+93*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+ 1/10*(5*A-5*B+9*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/30 *(15*A-35*B+39*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d
Time = 2.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-30 \sqrt {2} (7 A-11 B+15 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\frac {1}{2} (120 A-120 B+264 C+(225 A-205 B+393 C) \cos (c+d x)+24 (5 A-5 B+9 C) \cos (2 (c+d x))+75 A \cos (3 (c+d x))-95 B \cos (3 (c+d x))+147 C \cos (3 (c+d x))) \sqrt {1-\sec (c+d x)} \sec ^3(c+d x)\right ) \tan (c+d x)}{60 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^(3/2),x]
Output:
((-30*Sqrt[2]*(7*A - 11*B + 15*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]* Cos[(c + d*x)/2]^2*Sec[c + d*x] + ((120*A - 120*B + 264*C + (225*A - 205*B + 393*C)*Cos[c + d*x] + 24*(5*A - 5*B + 9*C)*Cos[2*(c + d*x)] + 75*A*Cos[ 3*(c + d*x)] - 95*B*Cos[3*(c + d*x)] + 147*C*Cos[3*(c + d*x)])*Sqrt[1 - Se c[c + d*x]]*Sec[c + d*x]^3)/2)*Tan[c + d*x])/(60*d*Sqrt[1 - Sec[c + d*x]]* (a*(1 + Sec[c + d*x]))^(3/2))
Time = 1.47 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4572, 27, 3042, 4509, 27, 3042, 4498, 27, 3042, 4489, 3042, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int -\frac {\sec ^3(c+d x) (2 a (A-3 B+3 C)-a (5 A-5 B+9 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^3(c+d x) (2 a (A-3 B+3 C)-a (5 A-5 B+9 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (2 a (A-3 B+3 C)-a (5 A-5 B+9 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^2 (5 A-5 B+9 C)-a^2 (15 A-35 B+39 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^2 (5 A-5 B+9 C)-a^2 (15 A-35 B+39 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^2 (5 A-5 B+9 C)-a^2 (15 A-35 B+39 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4498 |
| \(\displaystyle -\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^3 (15 A-35 B+39 C)-2 a^3 (45 A-65 B+93 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a (15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^3 (15 A-35 B+39 C)-2 a^3 (45 A-65 B+93 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a (15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^3 (15 A-35 B+39 C)-2 a^3 (45 A-65 B+93 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a (15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle -\frac {-\frac {-\frac {15 a^3 (7 A-11 B+15 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^3 (45 A-65 B+93 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-\frac {15 a^3 (7 A-11 B+15 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^3 (45 A-65 B+93 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle -\frac {-\frac {-\frac {-\frac {30 a^3 (7 A-11 B+15 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^3 (45 A-65 B+93 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {-\frac {-\frac {\frac {15 \sqrt {2} a^{5/2} (7 A-11 B+15 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^3 (45 A-65 B+93 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a (15 A-35 B+39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a (5 A-5 B+9 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
Input:
Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]
Output:
-1/2*((A - B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/ 2)) - ((-2*a*(5*A - 5*B + 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) - ((-2*a*(15*A - 35*B + 39*C)*Sqrt[a + a*Sec[c + d*x]]*Ta n[c + d*x])/(3*d) - ((15*Sqrt[2]*a^(5/2)*(7*A - 11*B + 15*C)*ArcTan[(Sqrt[ a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^3*(45*A - 6 5*B + 93*C)*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a))/(4*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[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_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* ((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int [Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) *Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a *B, 0] && !LtQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[d/(b*(m + n)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]
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 _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 3.65 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.64
| method | result | size |
| default | \(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-105 \cos \left (d x +c \right )^{2}-210 \cos \left (d x +c \right )-105\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (165 \cos \left (d x +c \right )^{2}+330 \cos \left (d x +c \right )+165\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-225 \cos \left (d x +c \right )^{2}-450 \cos \left (d x +c \right )-225\right ) \sqrt {2}\, C \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \left (150 \cos \left (d x +c \right )+120\right ) A +\left (-190 \cos \left (d x +c \right )^{2}-120 \cos \left (d x +c \right )+40\right ) B \tan \left (d x +c \right )+\left (294 \cos \left (d x +c \right )^{3}+216 \cos \left (d x +c \right )^{2}-24 \cos \left (d x +c \right )+24\right ) C \tan \left (d x +c \right ) \sec \left (d x +c \right )\right )}{60 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\) | \(375\) |
| parts | \(\frac {A \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{4}-\frac {7 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{4}+\frac {9 \csc \left (d x +c \right )}{4}-\frac {9 \cot \left (d x +c \right )}{4}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{2}}+\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-38 \cos \left (d x +c \right )^{2}-24 \cos \left (d x +c \right )+8\right ) \tan \left (d x +c \right )+\left (33 \cos \left (d x +c \right )^{2}+66 \cos \left (d x +c \right )+33\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{12 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}-\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-98 \cos \left (d x +c \right )^{3}-72 \cos \left (d x +c \right )^{2}+8 \cos \left (d x +c \right )-8\right ) \sec \left (d x +c \right ) \tan \left (d x +c \right )+\left (75 \cos \left (d x +c \right )^{2}+150 \cos \left (d x +c \right )+75\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{20 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\) | \(433\) |
Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, method=_RETURNVERBOSE)
Output:
1/60/d/a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)*((-105*c os(d*x+c)^2-210*cos(d*x+c)-105)*2^(1/2)*A*(-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))+(165*cos (d*x+c)^2+330*cos(d*x+c)+165)*2^(1/2)*B*(-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))+(-225*cos( d*x+c)^2-450*cos(d*x+c)-225)*2^(1/2)*C*(-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)* (150*cos(d*x+c)+120)*A+(-190*cos(d*x+c)^2-120*cos(d*x+c)+40)*B*tan(d*x+c)+ (294*cos(d*x+c)^3+216*cos(d*x+c)^2-24*cos(d*x+c)+24)*C*tan(d*x+c)*sec(d*x+ c))
Time = 0.13 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.33 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {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 ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (75 \, A - 95 \, B + 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 12 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac {15 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 11 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \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 ) + 2 \, {\left ({\left (75 \, A - 95 \, B + 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, A - 5 \, B + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 12 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3 /2),x, algorithm="fricas")
Output:
[-1/120*(15*sqrt(2)*((7*A - 11*B + 15*C)*cos(d*x + c)^4 + 2*(7*A - 11*B + 15*C)*cos(d*x + c)^3 + (7*A - 11*B + 15*C)*cos(d*x + c)^2)*sqrt(-a)*log(-( 2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*si n(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)) - 4*((75*A - 95*B + 147*C)*cos(d*x + c)^3 + 12*(5*A - 5*B + 9*C)*cos(d*x + c)^2 + 4*(5*B - 3*C)*cos(d*x + c) + 12*C)*sqrt((a*co s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2* d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2), 1/60*(15*sqrt(2)*((7*A - 11*B + 15*C)*cos(d*x + c)^4 + 2*(7*A - 11*B + 15*C)*cos(d*x + c)^3 + (7*A - 11*B + 15*C)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((75*A - 95*B + 147* C)*cos(d*x + c)^3 + 12*(5*A - 5*B + 9*C)*cos(d*x + c)^2 + 4*(5*B - 3*C)*co s(d*x + c) + 12*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/( a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* *(3/2),x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a*(sec( c + d*x) + 1))**(3/2), x)
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3 /2),x, algorithm="maxima")
Output:
Timed out
Time = 1.17 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (7 \, A - 11 \, B + 15 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left ({\left ({\left (\frac {15 \, \sqrt {2} {\left (A a^{3} - B a^{3} + C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (165 \, A a^{3} - 245 \, B a^{3} + 381 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, \sqrt {2} {\left (57 \, A a^{3} - 73 \, B a^{3} + 105 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, \sqrt {2} {\left (9 \, A a^{3} - 9 \, B a^{3} + 17 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{60 \, d} \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3 /2),x, algorithm="giac")
Output:
1/60*(15*sqrt(2)*(7*A - 11*B + 15*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c ) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a*sgn(cos(d*x + c))) - (((15*sqrt(2)*(A*a^3 - B*a^3 + C*a^3)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(cos (d*x + c))) - sqrt(2)*(165*A*a^3 - 245*B*a^3 + 381*C*a^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 5*sqrt(2)*(57*A*a^3 - 73*B*a^3 + 105*C*a ^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 15*sqrt(2)*(9*A*a^3 - 9*B*a^3 + 17*C*a^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)/((a*ta n(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)),x)
Output:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)), x)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{2}} \] Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**5)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**4)/(s ec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x) + 1)*se c(c + d*x)**3)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*a))/a**2