Integrand size = 45, antiderivative size = 179 \[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a} (8 A+6 B+5 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a (8 A+6 B+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}+\frac {a (6 B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {C \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d} \] Output:
1/8*a^(1/2)*(8*A+6*B+5*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2 ))/d+1/8*a*(8*A+6*B+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1 /2)+1/12*a*(6*B+C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/ 3*C*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d
Time = 1.19 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.91 \[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (3 (8 A+6 B+5 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (6 B+5 C) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+3 (8 A+6 B+5 C) \sqrt {(-1+\cos (c+d x)) \sec ^2(c+d x)}\right ) \tan (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(a*(3*(8*A + 6*B + 5*C)*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 2*(6*B + 5*C)*Sqr t[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 8*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 3*(8*A + 6*B + 5*C)*Sqrt[(-1 + Cos[c + d*x])*Sec[c + d*x]^2 ])*Tan[c + d*x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.92 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4576, 27, 3042, 4504, 3042, 4290, 3042, 4288, 222}
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.
\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4576 |
\(\displaystyle \frac {\int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+C)+a (6 B+C) \sec (c+d x))dx}{3 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+C)+a (6 B+C) \sec (c+d x))dx}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (2 A+C)+a (6 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 4504 |
\(\displaystyle \frac {\frac {3}{4} a (8 A+6 B+5 C) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{4} a (8 A+6 B+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 4290 |
\(\displaystyle \frac {\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 4288 |
\(\displaystyle \frac {\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\) |
Input:
Int[Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
(C*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d) + ((a^2 *(6*B + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (3*a*(8*A + 6*B + 5*C)*((Sqrt[a]*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a* Sec[c + d*x]])))/4)/(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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 1))) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, 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[((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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(426\) vs. \(2(153)=306\).
Time = 9.01 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.39
method | result | size |
default | \(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (-24 A \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-18 B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-15 C \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-24 A \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-18 B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-15 C \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+24 A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (18 \cos \left (d x +c \right )+12\right ) \sqrt {2}\, B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\left (15 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+8\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right )\right )}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(427\) |
parts | \(-\frac {A \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{2 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {B \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-3 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+\sin \left (d x +c \right ) \left (-3 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {C \sec \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-15 \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+15 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{4}+\sin \left (d x +c \right ) \left (-15 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-8\right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )\right )}{48 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(533\) |
Input:
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x,method=_RETURNVERBOSE)
Output:
1/48/d*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(3/2)/(cos(d*x+c)+1)/(-1/(cos(d *x+c)+1))^(1/2)*(-24*A*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(cos(d*x+ c)+1))^(1/2))*cos(d*x+c)^2-18*B*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/ (cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2-15*C*arctan(1/2*(-cot(d*x+c)+csc(d*x+c )+1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2-24*A*arctan(1/2*(-cot(d*x+c)+ csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2-18*B*arctan(1/2*(-co t(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2-15*C*arctan (1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2+24 *A*(-2/(cos(d*x+c)+1))^(1/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)+sin(d*x+c)*(18* cos(d*x+c)+12)*2^(1/2)*B*(-2/(cos(d*x+c)+1))^(1/2)+(15*cos(d*x+c)^2+10*cos (d*x+c)+8)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*tan(d*x+c))
Time = 0.28 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.59 \[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (3 \, {\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 8 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {3 \, {\left ({\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (3 \, {\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 8 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="fricas")
Output:
[1/96*(3*((8*A + 6*B + 5*C)*cos(d*x + c)^3 + (8*A + 6*B + 5*C)*cos(d*x + c )^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^ 2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d* x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3 *(8*A + 6*B + 5*C)*cos(d*x + c)^2 + 2*(6*B + 5*C)*cos(d*x + c) + 8*C)*sqrt ((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*co s(d*x + c)^3 + d*cos(d*x + c)^2), 1/48*(3*((8*A + 6*B + 5*C)*cos(d*x + c)^ 3 + (8*A + 6*B + 5*C)*cos(d*x + c)^2)*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt (cos(d*x + c))*sin(d*x + c))) + 2*(3*(8*A + 6*B + 5*C)*cos(d*x + c)^2 + 2* (6*B + 5*C)*cos(d*x + c) + 8*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si n(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)+C*sec( d*x+c)**2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 4002 vs. \(2 (153) = 306\).
Time = 0.45 (sec) , antiderivative size = 4002, normalized size of antiderivative = 22.36 \[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="maxima")
Output:
-1/96*(24*(4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d* x + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d* x + 2*c) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arcta n2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c) , cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)* log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(s in(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), co s(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log( 2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d *x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d* x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + ( cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*co s(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 4*(sq rt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x +...
Leaf count of result is larger than twice the leaf count of optimal. 1080 vs. \(2 (153) = 306\).
Time = 2.47 (sec) , antiderivative size = 1080, normalized size of antiderivative = 6.03 \[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d* x+c)^2),x, algorithm="giac")
Output:
1/48*(3*(8*A*sqrt(a)*sgn(cos(d*x + c)) + 6*B*sqrt(a)*sgn(cos(d*x + c)) + 5 *C*sqrt(a)*sgn(cos(d*x + 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))) - 3*(8*A*sqrt(a)*s gn(cos(d*x + c)) + 6*B*sqrt(a)*sgn(cos(d*x + c)) + 5*C*sqrt(a)*sgn(cos(d*x + 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))) + 4*(72*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1 /2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(3/2)*sgn(cos(d*x + c)) - 30*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^ 2 + a))^10*B*a^(3/2)*sgn(cos(d*x + c)) + 63*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*C*a^(3/2)*sgn(cos(d*x + c )) - 888*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2* c)^2 + a))^8*A*a^(5/2)*sgn(cos(d*x + c)) + 66*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*B*a^(5/2)*sgn(cos(d*x + c)) - 369*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2 *c)^2 + a))^8*C*a^(5/2)*sgn(cos(d*x + c)) + 3024*sqrt(2)*(sqrt(a)*tan(1/2* d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(7/2)*sgn(cos(d*x + c)) + 756*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(7/2)*sgn(cos(d*x + c)) + 1638*sqrt(2)*(sqrt(a)*tan(1 /2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C*a^(7/2)*sgn(cos( d*x + c)) - 1776*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2...
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:
int((a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
int((a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
\[ \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x)
Output:
sqrt(a)*(int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)* c + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*b + i nt(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*a)