Integrand size = 45, antiderivative size = 199 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a} (8 A+6 B+5 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a (6 B+C) \sin (c+d x)}{12 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (8 A+6 B+5 C) \sin (c+d x)}{8 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {C \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {5}{2}}(c+d x)} \] 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 ))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/12*a*(6*B+C)*sin(d*x+c)/d/cos(d*x +c)^(5/2)/(a+a*sec(d*x+c))^(1/2)+1/8*a*(8*A+6*B+5*C)*sin(d*x+c)/d/cos(d*x+ c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+1/3*C*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d/ cos(d*x+c)^(5/2)
Time = 0.72 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left (3 (8 A+6 B+5 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (6 B+5 C) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+3 (8 A+6 B+5 C) \sqrt {(-1+\cos (c+d x)) \sec ^2(c+d x)}\right ) \sin (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2 ))/Cos[c + d*x]^(3/2),x]
Output:
(a*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(3*(8*A + 6*B + 5*C)*ArcSin[Sqrt[ 1 - Sec[c + d*x]]] + 2*(6*B + 5*C)*Sqrt[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)*S qrt[(-1 + Cos[c + d*x])*Sec[c + d*x]^2])*Sin[c + d*x])/(24*d*Sqrt[1 - Sec[ c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.18 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 4753, 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.
\(\displaystyle \int \frac {\sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\cos (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4753 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx\) |
\(\Big \downarrow \) 4576 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+C)+a (6 B+C) \sec (c+d x))dx}{3 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (2 A+C)+a (6 B+C) \sec (c+d x))dx}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (2 A+C)+a (6 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 4504 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{4} a (8 A+6 B+5 C) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{4} a (8 A+6 B+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 4290 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 4288 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (6 B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a (8 A+6 B+5 C) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )}{6 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )\) |
Input:
Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Cos [c + d*x]^(3/2),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(5/2)*Sqrt[a + a*Se c[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]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(434\) vs. \(2(169)=338\).
Time = 3.11 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.19
method | result | size |
default | \(\frac {\left (24 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+18 B \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+15 C \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+24 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+18 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+15 C \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+24 A \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (18 \cos \left (d x +c \right )+12\right ) \sqrt {2}\, B \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right ) \left (15 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+8\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{48 d \cos \left (d x +c \right )^{\frac {5}{2}} \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(435\) |
Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2 ),x,method=_RETURNVERBOSE)
Output:
1/48/d*(24*A*cos(d*x+c)^3*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c) -csc(d*x+c)+1))+18*B*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc( d*x+c)+1))*cos(d*x+c)^3+15*C*cos(d*x+c)^3*arctan(1/2/(-1/(cos(d*x+c)+1))^( 1/2)*(cot(d*x+c)-csc(d*x+c)+1))+24*A*cos(d*x+c)^3*arctan(1/2*(cot(d*x+c)-c sc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+18*B*cos(d*x+c)^3*arctan(1/2*(cot( d*x+c)-csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+15*C*cos(d*x+c)^3*arctan(1 /2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+24*A*2^(1/2)*sin(d *x+c)*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2+sin(d*x+c)*cos(d*x+c)*(18*cos (d*x+c)+12)*2^(1/2)*B*(-2/(cos(d*x+c)+1))^(1/2)+sin(d*x+c)*(15*cos(d*x+c)^ 2+10*cos(d*x+c)+8)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2))*(a*(1+sec(d*x+c))) ^(1/2)/cos(d*x+c)^(5/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2)
Time = 0.21 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.30 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [\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 )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\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 )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 3 \, {\left ({\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (8 \, A + 6 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),x, algorithm="fricas")
Output:
[1/96*(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))*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((8*A + 6*B + 5*C)*cos(d*x + c)^4 + (8*A + 6*B + 5*C)*cos(d*x + c )^3)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos( d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3), 1/48*(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))*sqrt(cos( d*x + c))*sin(d*x + c) + 3*((8*A + 6*B + 5*C)*cos(d*x + c)^4 + (8*A + 6*B + 5*C)*cos(d*x + c)^3)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a )/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos( d*x + c) - 2*a)))/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)]
\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x +c)**(3/2),x)
Output:
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x) + C*sec(c + d*x)** 2)/cos(c + d*x)**(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 4002 vs. \(2 (169) = 338\).
Time = 0.44 (sec) , antiderivative size = 4002, normalized size of antiderivative = 20.11 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),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. \(2 (169) = 338\).
Time = 0.68 (sec) , antiderivative size = 1080, normalized size of antiderivative = 5.43 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c )^(3/2),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. \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c os(c + d*x)^(3/2),x)
Output:
int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/c os(c + d*x)^(3/2), x)
\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a \right ) \] Input:
int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2 ),x)
Output:
sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2)/c os(c + d*x)**2,x)*c + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x))/cos(c + d*x)**2,x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d *x)))/cos(c + d*x)**2,x)*a)