Integrand size = 45, antiderivative size = 235 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {(8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {C \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {(4 B-C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \]
1/4*(8*A-4*B+7*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/a^(1/2)-(A-B+C)*arctanh(1/2*sin(d*x+c)*a^(1 /2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1 /2)*sec(d*x+c)^(1/2)/d/a^(1/2)+1/2*C*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*se c(d*x+c))^(1/2)+1/4*(4*B-C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c)) ^(1/2)
Time = 1.17 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left ((4 B-C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-8 (A-B+C) \arcsin \left (\sqrt {\sec (c+d x)}\right )+4 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )-4 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+4 \sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+2 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+4 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}-C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(3/2)*Sqrt [a + a*Sec[c + d*x]]),x]
(Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*((4*B - C)*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 8*(A - B + C)*ArcSin[Sqrt[Sec[c + d*x]]] + 4*Sqrt[2]*A*ArcTan[(Sq rt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*B*ArcTan[(Sq rt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] + 4*Sqrt[2]*C*ArcTan[(Sq rt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] + 2*C*Sqrt[1 - Sec[c + d *x]]*Sec[c + d*x]^(3/2) + 4*B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] - C*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Sin[c + d*x])/(4*d*Sqrt[1 - S ec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.44 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4509, 27, 3042, 4511, 3042, 4288, 222, 4295, 219}
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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{3/2} \sqrt {a \sec (c+d x)+a}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{\sqrt {\sec (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (a (4 A+3 C)+a (4 B-C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (a (4 A+3 C)+a (4 B-C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a (4 A+3 C)+a (4 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((4 B-C) a^2+(8 A-4 B+7 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((4 B-C) a^2+(8 A-4 B+7 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left ((4 B-C) a^2+(8 A-4 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 4511 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a (8 A-4 B+7 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx-8 a^2 (A-B+C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a (8 A-4 B+7 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-8 a^2 (A-B+C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {-8 a^2 (A-B+C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a (8 A-4 B+7 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^{3/2} (8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-8 a^2 (A-B+C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {16 a^2 (A-B+C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{3/2} (8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^{3/2} (8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(5/2)*Sin[c + d*x]) /(2*d*Sqrt[a + a*Sec[c + d*x]]) + (((2*a^(3/2)*(8*A - 4*B + 7*C)*ArcSinh[( Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (8*Sqrt[2]*a^(3/2)*(A - B + C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[ a + a*Sec[c + d*x]])])/d)/(2*a) + (a*(4*B - C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a))
3.13.77.3.1 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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f)) Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; FreeQ[{a, b, d, e, f}, x] && 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[(-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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)/b Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Simp[B/b Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b , d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 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. \(602\) vs. \(2(196)=392\).
Time = 0.88 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.57
| method | result | size |
| default | \(\frac {\left (8 A \sqrt {2}\, \cos \left (d x +c \right )^{2} \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-8 B \sqrt {2}\, \cos \left (d x +c \right )^{2} \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+8 C \sqrt {2}\, \cos \left (d x +c \right )^{2} \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-8 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-8 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+4 B \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{2}+4 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+8 B \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}-7 C \cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-7 C \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-2 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}+4 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{8 d \left (1+\cos \left (d x +c \right )\right ) a \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{\frac {3}{2}}}\) | \(603\) |
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2 ),x,method=_RETURNVERBOSE)
1/8/d*(8*A*2^(1/2)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c ))/(-1/(1+cos(d*x+c)))^(1/2))-8*B*2^(1/2)*cos(d*x+c)^2*arctan(1/2*sin(d*x+ c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+8*C*2^(1/2)*cos(d*x+c )^2*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2) )-8*A*cos(d*x+c)^2*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(1+cos(d*x+c))/(- 1/(1+cos(d*x+c)))^(1/2))-8*A*cos(d*x+c)^2*arctan(1/2*(cos(d*x+c)+sin(d*x+c )+1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+4*B*arctan(1/2*(-cos(d*x+c) +sin(d*x+c)-1)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2+4*B* cos(d*x+c)^2*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+cos(d*x+c))/(-1/(1+co s(d*x+c)))^(1/2))+8*B*cos(d*x+c)*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2)-7*C* cos(d*x+c)^2*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(1+cos(d*x+c))/(-1/(1+c os(d*x+c)))^(1/2))-7*C*cos(d*x+c)^2*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/( 1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-2*C*cos(d*x+c)*sin(d*x+c)*(-1/(1+ cos(d*x+c)))^(1/2)+4*C*sin(d*x+c)*(-1/(1+cos(d*x+c)))^(1/2))*(a*(1+sec(d*x +c)))^(1/2)/(1+cos(d*x+c))/a/(-1/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(3/2)
Time = 0.45 (sec) , antiderivative size = 636, normalized size of antiderivative = 2.71 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, {\left ({\left (4 \, B - C\right )} \cos \left (d x + c\right ) + 2 \, 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 ) + {\left ({\left (8 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2}\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 ) + \frac {8 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{3} + {\left (A - B + C\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \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 )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{16 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}, \frac {8 \, \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{3} + {\left (A - B + C\right )} a \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (4 \, B - C\right )} \cos \left (d x + c\right ) + 2 \, 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 ) + {\left ({\left (8 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A - 4 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2}\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 )}{8 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}}\right ] \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c) )^(1/2),x, algorithm="fricas")
[1/16*(4*((4*B - C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A - 4*B + 7*C)*cos(d*x + c)^3 + (8*A - 4*B + 7*C)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt (a)*sqrt((a*cos(d*x + c) + a)/cos(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)) + 8*sqrt(2)*((A - B + C)*a*cos(d*x + c)^3 + (A - B + C)*a*cos(d* x + c)^2)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d *x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(co s(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos (d*x + c)^2), 1/8*(8*sqrt(2)*((A - B + C)*a*cos(d*x + c)^3 + (A - B + C)*a *cos(d*x + c)^2)*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d *x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c)) + 2*((4*B - C)*cos(d* x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*s in(d*x + c) + ((8*A - 4*B + 7*C)*cos(d*x + c)^3 + (8*A - 4*B + 7*C)*cos(d* x + c)^2)*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)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/(sqrt(a*(sec(c + d*x) + 1))*cos(c + d*x)**(3/2)), x)
Leaf count of result is larger than twice the leaf count of optimal. 3334 vs. \(2 (196) = 392\).
Time = 0.60 (sec) , antiderivative size = 3334, normalized size of antiderivative = 14.19 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c) )^(1/2),x, algorithm="maxima")
-1/16*(8*(sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin (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))) + 1) - sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d* x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*ar ctan2(sin(d*x + c), cos(d*x + 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) + 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) - 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)*s in(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + log(2*cos(1/2*arctan2(s in(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))*A/sqrt(a) + 4*(4*sqrt (2)*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*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)...
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\sqrt {a \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c) )^(1/2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(sqrt(a*sec(d*x + c) + a )*cos(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(a + a/cos (c + d*x))^(1/2)),x)