Integrand size = 43, antiderivative size = 213 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {(9 A-14 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {(7 A-2 B+8 C) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {(A-6 B) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {A \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}} \] Output:
-1/8*(9*A-14*B+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1 /2)/d+2^(1/2)*(A-B+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c ))^(1/2))/a^(1/2)/d+1/8*(7*A-2*B+8*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)- 1/12*(A-6*B)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*A*cos(d*x+ c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
Time = 0.53 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (-3 (9 A-14 B+8 C) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )+24 \sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) \left (3 (7 A-2 B+8 C)-2 (A-6 B) \cos (c+d x)+8 A \cos ^2(c+d x)\right ) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]
Output:
((-3*(9*A - 14*B + 8*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 24*Sqrt[2]*(A - B + C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(3*(7*A - 2* B + 8*C) - 2*(A - 6*B)*Cos[c + d*x] + 8*A*Cos[c + d*x]^2)*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x ])])
Time = 1.44 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.11, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 4574, 27, 3042, 4510, 27, 3042, 4510, 27, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 4574 |
\(\displaystyle \frac {\int -\frac {\cos ^2(c+d x) (a (A-6 B)-a (5 A+6 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos ^2(c+d x) (a (A-6 B)-a (5 A+6 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{6 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-6 B)-a (5 A+6 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{6 a}\) |
\(\Big \downarrow \) 4510 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {3 \cos (c+d x) \left (a^2 (7 A-2 B+8 C)-a^2 (A-6 B) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \int \frac {\cos (c+d x) \left (a^2 (7 A-2 B+8 C)-a^2 (A-6 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}}{6 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \int \frac {a^2 (7 A-2 B+8 C)-a^2 (A-6 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}}{6 a}\) |
\(\Big \downarrow \) 4510 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {\int -\frac {a^3 (9 A-14 B+8 C)-a^3 (7 A-2 B+8 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (9 A-14 B+8 C)-a^3 (7 A-2 B+8 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (9 A-14 B+8 C)-a^3 (7 A-2 B+8 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 4408 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (9 A-14 B+8 C) \int \sqrt {\sec (c+d x) a+a}dx-16 a^3 (A-B+C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (9 A-14 B+8 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-16 a^3 (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 4261 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-16 a^3 (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^3 (9 A-14 B+8 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (9 A-14 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-16 a^3 (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {32 a^3 (A-B+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{5/2} (9 A-14 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {A \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a (A-6 B) \sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {3 \left (\frac {a^2 (7 A-2 B+8 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (9 A-14 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {16 \sqrt {2} a^{5/2} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\right )}{4 a}}{6 a}\) |
Input:
Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Se c[c + d*x]],x]
Output:
(A*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - ((a*(A - 6*B)*Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) - (3*(-1/2* ((2*a^(5/2)*(9*A - 14*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Se c[c + d*x]]])/d - (16*Sqrt[2]*a^(5/2)*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (a^2*(7*A - 2*B + 8*C)*S in[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/(4*a))/(6*a)
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*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_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), 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 - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(575\) vs. \(2(184)=368\).
Time = 30.15 (sec) , antiderivative size = 576, normalized size of antiderivative = 2.70
method | result | size |
default | \(\frac {\left (\left (27 \cos \left (d x +c \right )+27\right ) A \sqrt {-\frac {\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 (-42 \cos \left (d x +c \right )-42\right ) B \sqrt {-\frac {\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 (24 \cos \left (d x +c \right )+24\right ) C \sqrt {-\frac {\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 (24 \cos \left (d x +c \right )+24\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 (-24 \cos \left (d x +c \right )-24\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 (24 \cos \left (d x +c \right )+24\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 ) \cos \left (d x +c \right ) \left (8 \cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )+21\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (12 \cos \left (d x +c \right )-6\right ) B +24 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 )}}{24 d a \left (\cos \left (d x +c \right )+1\right )}\) | \(576\) |
Input:
int(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, method=_RETURNVERBOSE)
Output:
1/24/d/a*((27*cos(d*x+c)+27)*A*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh( 2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot (d*x+c)^2-1)^(1/2))+(-42*cos(d*x+c)-42)*B*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/ 2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot (d*x+c)+cot(d*x+c)^2-1)^(1/2))+(24*cos(d*x+c)+24)*C*(-cos(d*x+c)/(cos(d*x+ c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc( d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2))+(24*cos(d*x+c)+24)*2^(1/2)*A*(-co s(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))+(-24*cos(d*x+c)-24)*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))+ (24*cos(d*x+c)+24)*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)*cos(d*x+c) *(8*cos(d*x+c)^2-2*cos(d*x+c)+21)*A+sin(d*x+c)*cos(d*x+c)*(12*cos(d*x+c)-6 )*B+24*C*cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)
Time = 8.20 (sec) , antiderivative size = 562, normalized size of antiderivative = 2.64 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1 /2),x, algorithm="fricas")
Output:
[1/48*(24*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*a)*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)/(cos(d*x + c)^ 2 + 2*cos(d*x + c) + 1)) - 3*((9*A - 14*B + 8*C)*cos(d*x + c) + 9*A - 14*B + 8*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)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d *x + c) + 1)) + 2*(8*A*cos(d*x + c)^3 - 2*(A - 6*B)*cos(d*x + c)^2 + 3*(7* A - 2*B + 8*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d *x + c))/(a*d*cos(d*x + c) + a*d), 1/24*(3*((9*A - 14*B + 8*C)*cos(d*x + c ) + 9*A - 14*B + 8*C)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c ))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (8*A*cos(d*x + c)^3 - 2*(A - 6*B )*cos(d*x + c)^2 + 3*(7*A - 2*B + 8*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 24*sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + C)*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)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)]
\[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \] Input:
integrate(cos(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* *(1/2),x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**3/sqrt(a*( sec(c + d*x) + 1)), x)
\[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1 /2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^3/sqrt(a*se c(d*x + c) + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 1076 vs. \(2 (184) = 368\).
Time = 1.09 (sec) , antiderivative size = 1076, normalized size of antiderivative = 5.05 \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1 /2),x, algorithm="giac")
Output:
-1/48*(24*sqrt(2)*(A - B + 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)*sgn(cos(d*x + c))) + 3*(9*A - 1 4*B + 8*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))^2 - a*(2*sqrt(2) + 3)))/(sqrt(-a)*sgn(cos(d*x + c))) - 3*(9 *A - 14*B + 8*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))^2 + a*(2*sqrt(2) - 3)))/(sqrt(-a)*sgn(cos(d*x + c))) + 4*sqrt(2)*(165*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/ 2*c)^2 + a))^10*A*sqrt(-a) - 102*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a* tan(1/2*d*x + 1/2*c)^2 + a))^10*B*sqrt(-a) + 72*(sqrt(-a)*tan(1/2*d*x + 1/ 2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C*sqrt(-a) - 1323*(sqrt(-a) *tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*sqrt(-a)* a + 954*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*sqrt(-a)*a - 888*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2* d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a + 3906*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^2 - 2268*(sqrt(-a)*t an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^ 2 + 3024*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^2 - 2118*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan( 1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^3 + 1044*(sqrt(-a)*tan(1/2*d*x + 1 /2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^3 - 1776*(s...
Timed out. \[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((cos(c + d*x)^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2),x)
Output:
int((cos(c + d*x)^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2), x)
\[ \int \frac {\cos ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) a \right )}{a} \] Input:
int(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x)**2)/(se c(c + d*x) + 1),x)*c + int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)**3*sec(c + d*x))/(sec(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x) + 1)*cos(c + d*x)* *3)/(sec(c + d*x) + 1),x)*a))/a