Integrand size = 35, antiderivative size = 227 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (200 A+163 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^3 (200 A+163 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^3 (104 A+95 B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (8 A+11 B) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {a B \sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d} \] Output:
1/64*a^(5/2)*(200*A+163*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/ 2))/d+1/64*a^3*(200*A+163*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c) )^(1/2)+1/96*a^3*(104*A+95*B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c ))^(1/2)+1/24*a^2*(8*A+11*B)*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(1/2)*sin(d *x+c)/d+1/4*a*B*sec(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d
Time = 0.95 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.84 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^3 \left ((600 A+489 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 (136 A+163 B) \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 (8 A+23 B) \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+48 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x)+3 (200 A+163 B) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{192 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x ]),x]
Output:
(a^3*((600*A + 489*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 2*(136*A + 163*B)*S qrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 8*(8*A + 23*B)*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) + 48*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(7/2) + 3*(200*A + 163*B)*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Tan[c + d* x])/(192*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.26 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4506, 27, 3042, 4506, 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.
| \(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} (a (8 A+3 B)+a (8 A+11 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^{3/2} (a (8 A+3 B)+a (8 A+11 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (8 A+3 B)+a (8 A+11 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{3} \int \frac {1}{2} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (24 A+17 B) a^2+(104 A+95 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a} \left (3 (24 A+17 B) a^2+(104 A+95 B) \sec (c+d x) a^2\right )dx+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 (24 A+17 B) a^2+(104 A+95 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (104 A+95 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (104 A+95 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4290 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (104 A+95 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^3 (104 A+95 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle \frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} a^2 (200 A+163 B) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a^3 (104 A+95 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle \frac {1}{8} \left (\frac {a^2 (8 A+11 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {1}{6} \left (\frac {a^3 (104 A+95 B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}+\frac {3}{4} a^2 (200 A+163 B) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d}\) |
Input:
Int[Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
Output:
(a*B*Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*d) + ( (a^2*(8*A + 11*B)*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x] )/(3*d) + ((a^3*(104*A + 95*B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[ a + a*Sec[c + d*x]]) + (3*a^2*(200*A + 163*B)*((Sqrt[a]*ArcSinh[(Sqrt[a]*T an[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)/8
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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Time = 4.48 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.48
| method | result | size |
| default | \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (600 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+489 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+600 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+489 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (600 \cos \left (d x +c \right )^{2}+272 \cos \left (d x +c \right )+64\right ) \sqrt {2}\, A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )+\left (489 \cos \left (d x +c \right )^{3}+326 \cos \left (d x +c \right )^{2}+184 \cos \left (d x +c \right )+48\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )\right )}{384 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\) | \(336\) |
| parts | \(\frac {A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (75 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+75 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (75 \cos \left (d x +c \right )^{2}+34 \cos \left (d x +c \right )+8\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}+\frac {B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (489 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+489 \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{3}+\left (489 \cos \left (d x +c \right )^{3}+326 \cos \left (d x +c \right )^{2}+184 \cos \left (d x +c \right )+48\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )}{384 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\) | \(380\) |
Input:
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x,method=_RET URNVERBOSE)
Output:
1/384/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(3/2)/(1+cos(d*x+c))/(-1/( 1+cos(d*x+c)))^(1/2)*(600*A*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-csc(d*x+c) -1)/(-1/(1+cos(d*x+c)))^(1/2))+489*B*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-c sc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))+600*A*cos(d*x+c)^2*arctan(1/2*(cot (d*x+c)-csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+489*B*cos(d*x+c)^2*arctan (1/2*(cot(d*x+c)-csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+(600*cos(d*x+c)^ 2+272*cos(d*x+c)+64)*2^(1/2)*A*(-2/(1+cos(d*x+c)))^(1/2)*tan(d*x+c)+(489*c os(d*x+c)^3+326*cos(d*x+c)^2+184*cos(d*x+c)+48)*2^(1/2)*B*(-2/(1+cos(d*x+c )))^(1/2)*tan(d*x+c)*sec(d*x+c))
Time = 0.21 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.27 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{3}\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 (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (136 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, A + 23 \, B\right )} a^{2} \cos \left (d x + c\right ) + 48 \, B a^{2}\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 )}}}{768 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {3 \, {\left ({\left (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{3}\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 (200 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (136 \, A + 163 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, A + 23 \, B\right )} a^{2} \cos \left (d x + c\right ) + 48 \, B a^{2}\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 )}}}{384 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algo rithm="fricas")
Output:
[1/768*(3*((200*A + 163*B)*a^2*cos(d*x + c)^4 + (200*A + 163*B)*a^2*cos(d* x + c)^3)*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))*s in(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*(200*A + 163*B)*a^2*cos(d*x + c)^3 + 2*(136*A + 163*B)*a^2*cos(d*x + c)^2 + 8*(8*A + 23*B)*a^2*cos(d*x + c) + 48*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4 + d*c os(d*x + c)^3), 1/384*(3*((200*A + 163*B)*a^2*cos(d*x + c)^4 + (200*A + 16 3*B)*a^2*cos(d*x + c)^3)*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*(200*A + 163*B)*a^2*cos(d*x + c)^3 + 2*(136*A + 16 3*B)*a^2*cos(d*x + c)^2 + 8*(8*A + 23*B)*a^2*cos(d*x + c) + 48*B*a^2)*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)^4 + d*cos(d*x + c)^3)]
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(3/2)*(a+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 7331 vs. \(2 (195) = 390\).
Time = 0.67 (sec) , antiderivative size = 7331, normalized size of antiderivative = 32.30 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algo rithm="maxima")
Output:
1/768*(8*(300*sqrt(2)*a^2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d* x + 3/2*c)))*sin(6*d*x + 6*c) - 28*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) + 28*s qrt(2)*a^2*sin(3/2*d*x + 3/2*c) - 28*(sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - s qrt(2)*a^2*sin(3/2*d*x + 3/2*c))*cos(6*d*x + 6*c) - 300*(sqrt(2)*a^2*sin(6 *d*x + 6*c) + 3*sqrt(2)*a^2*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* d*x + 3/2*c))) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3 /2*d*x + 3/2*c))))*cos(11/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/ 2*c))) - 12*(7*sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2*d* x + 3/2*c) - 114*sqrt(2)*a^2*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2 *d*x + 3/2*c))) + 114*sqrt(2)*a^2*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), co s(3/2*d*x + 3/2*c))) + 75*sqrt(2)*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c) , cos(3/2*d*x + 3/2*c))))*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d* x + 3/2*c))) - 456*(sqrt(2)*a^2*sin(6*d*x + 6*c) + 3*sqrt(2)*a^2*sin(4/3*a rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin(3 /2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 456*(sqrt(2)*a^2*sin(6*d*x + 6*c ) + 3*sqrt(2)*a^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* c))))*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*(7 *sqrt(2)*a^2*sin(9/2*d*x + 9/2*c) - 7*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 7 5*sqrt(2)*a^2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) )*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 75*(a^...
Leaf count of result is larger than twice the leaf count of optimal. 942 vs. \(2 (195) = 390\).
Time = 3.88 (sec) , antiderivative size = 942, normalized size of antiderivative = 4.15 \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algo rithm="giac")
Output:
1/384*(3*(200*A*a^(5/2)*sgn(cos(d*x + c)) + 163*B*a^(5/2)*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*(200*A*a^(5/2)*sgn(cos(d*x + c)) + 163*B* a^(5/2)*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*sqrt(2)*(600*(sqr t(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(7/ 2)*sgn(cos(d*x + c)) + 489*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2* d*x + 1/2*c)^2 + a))^14*B*a^(7/2)*sgn(cos(d*x + c)) - 12600*(sqrt(a)*tan(1 /2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*a^(9/2)*sgn(cos (d*x + c)) - 10269*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/ 2*c)^2 + a))^12*B*a^(9/2)*sgn(cos(d*x + c)) + 103992*(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^(11/2)*sgn(cos(d*x + c)) + 69885*(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^(11/2)*sgn(cos(d*x + c)) - 339864*(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^(13/2)*sgn(cos(d*x + c)) - 259233*(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^(13/2)*sgn(cos(d*x + c)) + 262920*(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^(15/2)*sgn(cos(d*x + c)) + 20997 9*(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^(15/2)*sgn(cos(d*x + c)) - 73640*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt...
Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(3/2) ,x)
Output:
int((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(3/2) , x)
\[ \int \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) a +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}d x \right ) b +2 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) a +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x \right ) a \right ) \] Input:
int(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)
Output:
sqrt(a)*a**2*(int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)** 4,x)*b + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)* a + 2*int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x)*b + 2*int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*a + in t(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*b + int(sqr t(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*a)