Integrand size = 45, antiderivative size = 242 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(2 B-3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{3/2} d}+\frac {(A-5 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+3 C) \sin (c+d x)}{2 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \] Output:
(2*B-3*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)/a^(3/2)/d+1/4*(A-5*B+9*C)*arctanh(1/2*a^(1/2)*sec(d*x +c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec( d*x+c)^(1/2)*2^(1/2)/a^(3/2)/d-1/2*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/( a+a*sec(d*x+c))^(3/2)+1/2*(A-B+3*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*s ec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(582\) vs. \(2(242)=484\).
Time = 3.81 (sec) , antiderivative size = 582, normalized size of antiderivative = 2.40 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (8 (A-B+3 C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+8 (A-5 B+9 C) \arcsin \left (\sqrt {\sec (c+d x)}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+2 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-2 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+6 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+4 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-\sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)+5 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-9 \sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-\sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)+5 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-9 \sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)\right )}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)),x]
Output:
(Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(8*(A - B + 3*C)*ArcSin[Sqrt[1 - Se c[c + d*x]]]*Cos[(c + d*x)/2]^3*Sec[c + d*x]^2*Sin[(c + d*x)/2] + 8*(A - 5 *B + 9*C)*ArcSin[Sqrt[Sec[c + d*x]]]*Cos[(c + d*x)/2]^3*Sec[c + d*x]^2*Sin [(c + d*x)/2] + 2*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] - 2*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 6*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 4*C*Sqrt[1 - Sec[c + d* x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] - Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + 5*Sqrt[2]*B*ArcTan[(Sqrt[ 2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 9*Sqrt[2]*C* ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec [c + d*x]*Tan[c + d*x] + 5*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/S qrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] - 9*Sqrt[2]*C*ArcTan[(Sqr t[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] ))/(4*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
Time = 1.53 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.95, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 4753, 3042, 4572, 27, 3042, 4509, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec (c+d x)^2}{\cos (c+d x)^{3/2} (a \sec (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 4753 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )}{(\sec (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (a (A+3 B-3 C)+2 a (A-B+3 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (a (A+3 B-3 C)+2 a (A-B+3 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a (A+3 B-3 C)+2 a (A-B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((A-B+3 C) a^2+2 (2 B-3 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left ((A-B+3 C) a^2+2 (2 B-3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4511 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (A-5 B+9 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx+2 a (2 B-3 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (A-5 B+9 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+2 a (2 B-3 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (A-5 B+9 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a (2 B-3 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (A-5 B+9 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {4 a^{3/2} (2 B-3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {4 a^{3/2} (2 B-3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^2 (A-5 B+9 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\sqrt {2} a^{3/2} (A-5 B+9 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {4 a^{3/2} (2 B-3 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a (A-B+3 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\) |
Input:
Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + a*Sec [c + d*x])^(3/2)),x]
Output:
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A - B + C)*Sec[c + d*x]^(5/2 )*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (((4*a^(3/2)*(2*B - 3*C)* ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (Sqrt[2]*a^( 3/2)*(A - 5*B + 9*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sq rt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (2*a*(A - B + 3*C)*Sec[c + d*x]^( 3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a^2))
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[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
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. \(1921\) vs. \(2(203)=406\).
Time = 4.74 (sec) , antiderivative size = 1922, normalized size of antiderivative = 7.94
Input:
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))^(3/2 ),x,method=_RETURNVERBOSE)
Output:
1/8*2^(1/2)/a*(-8*A/d*((4*cos(1/2*d*x+1/2*c)^2-2)*sin(1/2*d*x+1/2*c)+(8*co s(1/2*d*x+1/2*c)^4-8*cos(1/2*d*x+1/2*c)^2+2)*ln(-cot(1/2*d*x+1/2*c)+csc(1/ 2*d*x+1/2*c)+1)+(1+4*cos(1/2*d*x+1/2*c)^4-4*cos(1/2*d*x+1/2*c)^2)*2^(1/2)* arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)+1))+(1+4*cos(1/ 2*d*x+1/2*c)^4-4*cos(1/2*d*x+1/2*c)^2)*2^(1/2)*arctanh(1/2*2^(1/2)*(cot(1/ 2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)-1))+(-8*cos(1/2*d*x+1/2*c)^4+8*cos(1/2*d*x +1/2*c)^2-2)*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1))*cos(1/2*d*x+1/2 *c)/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d *x+1/2*c)^2-1)^(5/2)+4*B/d*((4*cos(1/2*d*x+1/2*c)^2-2)*sin(1/2*d*x+1/2*c)+ (8*cos(1/2*d*x+1/2*c)^4-8*cos(1/2*d*x+1/2*c)^2+2)*ln(-cot(1/2*d*x+1/2*c)+c sc(1/2*d*x+1/2*c)+1)+(1+4*cos(1/2*d*x+1/2*c)^4-4*cos(1/2*d*x+1/2*c)^2)*2^( 1/2)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)+1))+(1+4*c os(1/2*d*x+1/2*c)^4-4*cos(1/2*d*x+1/2*c)^2)*2^(1/2)*arctanh(1/2*2^(1/2)*(c ot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)-1))+(-8*cos(1/2*d*x+1/2*c)^4+8*cos(1/ 2*d*x+1/2*c)^2-2)*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1))*cos(1/2*d* x+1/2*c)/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos( 1/2*d*x+1/2*c)^2-1)^(5/2)+2*A/d/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+ 1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(5/2)*((12*cos(1/2*d*x+1/2*c)^4 -8*cos(1/2*d*x+1/2*c)^2+1)*tan(1/2*d*x+1/2*c)+ln(-cot(1/2*d*x+1/2*c)+csc(1 /2*d*x+1/2*c)+1)*(36*cos(1/2*d*x+1/2*c)^5-36*cos(1/2*d*x+1/2*c)^3+9*cos...
Time = 0.17 (sec) , antiderivative size = 736, normalized size of antiderivative = 3.04 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
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) )^(3/2),x, algorithm="fricas")
Output:
[1/8*(sqrt(2)*((A - 5*B + 9*C)*cos(d*x + c)^3 + 2*(A - 5*B + 9*C)*cos(d*x + c)^2 + (A - 5*B + 9*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2* sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c)) *sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((A - B + 3*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) - 2*((2*B - 3*C)*cos(d*x + c)^3 + 2*(2*B - 3*C)*cos(d*x + c)^2 + (2*B - 3*C)*cos(d*x + c))*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)/(c os(d*x + c)^3 + cos(d*x + c)^2)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c)), -1/4*(sqrt(2)*((A - 5*B + 9*C)*cos(d*x + c)^ 3 + 2*(A - 5*B + 9*C)*cos(d*x + c)^2 + (A - 5*B + 9*C)*cos(d*x + c))*sqrt( -a)*arctan(1/2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq rt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c) + a)) - 2*((A - B + 3*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) - 2*((2*B - 3*C)*cos(d*x + c)^3 + 2*(2*B - 3*C)*cos(d*x + c )^2 + (2*B - 3*C)*cos(d*x + c))*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^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
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))**(3/2),x)
Output:
Timed out
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
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) )^(3/2),x, algorithm="maxima")
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (203) = 406\).
Time = 165.43 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.77 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (A \sqrt {a} - 5 \, B \sqrt {a} + 9 \, C \sqrt {a}\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 )}^{2}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {4 \, {\left (2 \, B - 3 \, C\right )} \log \left ({\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 )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {4 \, {\left (2 \, B - 3 \, C\right )} \log \left ({\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 )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {16 \, \sqrt {2} {\left (3 \, {\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 )}^{2} C \sqrt {a} - C a^{\frac {3}{2}}\right )}}{{\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 )}^{4} - 6 \, {\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 )}^{2} a + a^{2}\right )} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{8 \, d} \] Input:
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) )^(3/2),x, algorithm="giac")
Output:
-1/8*(sqrt(2)*(A*sqrt(a) - 5*B*sqrt(a) + 9*C*sqrt(a))*log((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*sgn(cos(d*x + c))) - 4*(2*B - 3*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)))/(a^(3/2)*sgn(cos(d*x + c))) + 4*(2*B - 3*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)))/(a^(3/2)*sgn(cos(d*x + c))) - 2 *(sqrt(2)*A*a*sgn(cos(d*x + c)) - sqrt(2)*B*a*sgn(cos(d*x + c)) + sqrt(2)* C*a*sgn(cos(d*x + c)))*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/ 2*c)/a^3 - 16*sqrt(2)*(3*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d* x + 1/2*c)^2 + a))^2*C*sqrt(a) - C*a^(3/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c ) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c ) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)*a*sgn(cos(d*x + c))))/d
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:
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))^(3/2)),x)
Output:
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))^(3/2)), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{2}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )+\cos \left (d x +c \right )^{2}}d x \right ) a \right )}{a^{2}} \] Input:
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))^(3/2 ),x)
Output:
(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x)**2)/ (cos(c + d*x)**2*sec(c + d*x)**2 + 2*cos(c + d*x)**2*sec(c + d*x) + cos(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*sec(c + d*x)**2 + 2*cos(c + d*x)**2*sec(c + d*x) + c os(c + d*x)**2),x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x)))/(co s(c + d*x)**2*sec(c + d*x)**2 + 2*cos(c + d*x)**2*sec(c + d*x) + cos(c + d *x)**2),x)*a))/a**2