Integrand size = 42, antiderivative size = 243 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 (111 B-143 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}-\frac {2 (3 B-19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}+\frac {2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt {a+a \sec (c+d x)}}+\frac {2 (93 B-29 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 a d} \]
(B-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 )/d/a^(1/2)-4/315*(111*B-143*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/105* (3*B-19*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*(9*B-C)*s ec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/9*C*sec(d*x+c)^4*tan(d*x +c)/d/(a+a*sec(d*x+c))^(1/2)+2/315*(93*B-29*C)*(a+a*sec(d*x+c))^(1/2)*tan( d*x+c)/a/d
Time = 0.85 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (315 \sqrt {2} (B-C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\frac {1}{4} (-423 B+1279 C+(918 B-214 C) \cos (c+d x)-8 (69 B-157 C) \cos (2 (c+d x))+186 B \cos (3 (c+d x))-58 C \cos (3 (c+d x))-129 B \cos (4 (c+d x))+257 C \cos (4 (c+d x))) \sqrt {1-\sec (c+d x)} \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
((315*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + ((-423*B + 1279*C + (918*B - 214*C)*Cos[c + d*x] - 8*(69*B - 157*C)*Cos[2*(c + d*x)] + 186*B*Cos[3*(c + d*x)] - 58*C*Cos[3*(c + d*x)] - 129*B*Cos[4*(c + d*x)] + 257*C*Cos[4*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^4)/4)*Tan[c + d*x])/(315*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.85 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.16, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4560, 3042, 4509, 27, 3042, 4509, 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.
\(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4560 |
\(\displaystyle \int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{\sqrt {a \sec (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4509 |
\(\displaystyle \frac {2 \int \frac {\sec ^4(c+d x) (8 a C+a (9 B-C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (8 a C+a (9 B-C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (8 a C+a (9 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4509 |
\(\displaystyle \frac {\frac {2 \int \frac {3 \sec ^3(c+d x) \left (2 a^2 (9 B-C)-a^2 (3 B-19 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 \int \frac {\sec ^3(c+d x) \left (2 a^2 (9 B-C)-a^2 (3 B-19 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (2 a^2 (9 B-C)-a^2 (3 B-19 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4509 |
\(\displaystyle \frac {\frac {3 \left (\frac {2 \int -\frac {\sec ^2(c+d x) \left (4 a^3 (3 B-19 C)-a^3 (93 B-29 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {\sec ^2(c+d x) \left (4 a^3 (3 B-19 C)-a^3 (93 B-29 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a^3 (3 B-19 C)-a^3 (93 B-29 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4498 |
\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (a^4 (93 B-29 C)-2 a^4 (111 B-143 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {\int \frac {\sec (c+d x) \left (a^4 (93 B-29 C)-2 a^4 (111 B-143 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a^4 (93 B-29 C)-2 a^4 (111 B-143 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4489 |
\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {315 a^4 (B-C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {315 a^4 (B-C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {-\frac {630 a^4 (B-C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {3 \left (-\frac {2 a^2 (3 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{5 d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a^2 (93 B-29 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}-\frac {\frac {315 \sqrt {2} a^{7/2} (B-C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^4 (111 B-143 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}\right )}{7 a}+\frac {2 a (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}\) |
(2*C*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) + ((2*a*( 9*B - C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (3* ((-2*a^2*(3*B - 19*C)*Sec[c + d*x]^2*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) - ((-2*a^2*(93*B - 29*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3 *d) - ((315*Sqrt[2]*a^(7/2)*(B - C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2] *Sqrt[a + a*Sec[c + d*x]])])/d - (4*a^4*(111*B - 143*C)*Tan[c + d*x])/(d*S qrt[a + a*Sec[c + d*x]]))/(3*a))/(5*a)))/(7*a))/(9*a)
3.4.84.3.1 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[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_.))^(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]
Time = 0.98 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.67
method | result | size |
default | \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (315 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {9}{2}}-315 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {9}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-552 B \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+766 C \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+1224 B \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-1872 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-1512 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+3276 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+840 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-1680 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+630 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{315 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{4}}\) | \(405\) |
parts | \(\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (105 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {7}{2}}-184 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+224 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-280 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}\right )}{105 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3}}-\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (315 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {9}{2}}-766 \left (1-\cos \left (d x +c \right )\right )^{9} \csc \left (d x +c \right )^{9}+1872 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}-3276 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+1680 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+630 \cot \left (d x +c \right )-630 \csc \left (d x +c \right )\right )}{315 d a \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{4}}\) | \(438\) |
int(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,me thod=_RETURNVERBOSE)
1/315/d/a*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(315*B*ln(csc(d*x +c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2* csc(d*x+c)^2-1)^(9/2)-315*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(9/2)*ln(csc (d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-552*B*(1-cos(d *x+c))^9*csc(d*x+c)^9+766*C*(1-cos(d*x+c))^9*csc(d*x+c)^9+1224*B*(1-cos(d* x+c))^7*csc(d*x+c)^7-1872*C*(1-cos(d*x+c))^7*csc(d*x+c)^7-1512*B*(1-cos(d* x+c))^5*csc(d*x+c)^5+3276*C*(1-cos(d*x+c))^5*csc(d*x+c)^5+840*B*(1-cos(d*x +c))^3*csc(d*x+c)^3-1680*C*(1-cos(d*x+c))^3*csc(d*x+c)^3+630*C*(-cot(d*x+c )+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^4
Time = 0.30 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.91 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {315 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{5} + {\left (B - C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (129 \, B - 257 \, C\right )} \cos \left (d x + c\right )^{4} - {\left (93 \, B - 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (9 \, B - C\right )} \cos \left (d x + c\right ) - 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{630 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}, -\frac {2 \, {\left ({\left (129 \, B - 257 \, C\right )} \cos \left (d x + c\right )^{4} - {\left (93 \, B - 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (9 \, B - C\right )} \cos \left (d x + c\right ) - 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac {315 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right )^{5} + {\left (B - C\right )} a \cos \left (d x + c\right )^{4}\right )} \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 )}{\sqrt {a}}}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}\right ] \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2 ),x, algorithm="fricas")
[-1/630*(315*sqrt(2)*((B - C)*a*cos(d*x + c)^5 + (B - C)*a*cos(d*x + c)^4) *sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1 /a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c) - 1)/(co s(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((129*B - 257*C)*cos(d*x + c)^4 - (93*B - 29*C)*cos(d*x + c)^3 + 3*(3*B - 19*C)*cos(d*x + c)^2 - 5*(9*B - C) *cos(d*x + c) - 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) )/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4), -1/315*(2*((129*B - 257*C)*co s(d*x + c)^4 - (93*B - 29*C)*cos(d*x + c)^3 + 3*(3*B - 19*C)*cos(d*x + c)^ 2 - 5*(9*B - C)*cos(d*x + c) - 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c ))*sin(d*x + c) + 315*sqrt(2)*((B - C)*a*cos(d*x + c)^5 + (B - C)*a*cos(d* x + c)^4)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c )^4)]
\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{5}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{4}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2 ),x, algorithm="maxima")
Time = 1.43 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.32 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {315 \, {\left (\sqrt {2} B - \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, {\left (\frac {315 \, \sqrt {2} C a^{4}}{\mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + {\left (420 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 840 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (756 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 1638 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (612 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 936 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - {\left (276 \, \sqrt {2} B a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 383 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{315 \, d} \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2 ),x, algorithm="giac")
-1/315*(315*(sqrt(2)*B - sqrt(2)*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)*sgn(cos(d*x + c))) - 2* (315*sqrt(2)*C*a^4/sgn(cos(d*x + c)) + (420*sqrt(2)*B*a^4*sgn(cos(d*x + c) ) - 840*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (756*sqrt(2)*B*a^4*sgn(cos(d*x + c)) - 1638*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (612*sqrt(2)*B*a^4*sgn(cos(d *x + c)) - 936*sqrt(2)*C*a^4*sgn(cos(d*x + c)) - (276*sqrt(2)*B*a^4*sgn(co s(d*x + c)) - 383*sqrt(2)*C*a^4*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2) *tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*t an(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d
Timed out. \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]