Integrand size = 23, antiderivative size = 92 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=\frac {5 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}+\frac {a^3 \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}}+\frac {a^2 \sqrt {a+a \cos (c+d x)} \tan (c+d x)}{d} \] Output:
5*a^(5/2)*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/d+a^3*sin(d*x +c)/d/(a+a*cos(d*x+c))^(1/2)+a^2*(a+a*cos(d*x+c))^(1/2)*tan(d*x+c)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 9.25 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.36 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=\frac {(a (1+\cos (c+d x)))^{5/2} \left (-\frac {35}{8} (2080+3131 \cos (c+d x)+728 \cos (2 (c+d x))+61 \cos (3 (c+d x))) \csc ^6\left (\frac {1}{2} (c+d x)\right )+\frac {105 \text {arctanh}\left (\sqrt {1-\cos (c+d x)}\right ) (1767+1252 \cos (c+d x)+872 \cos (2 (c+d x))+108 \cos (3 (c+d x))+\cos (4 (c+d x))) \csc ^6\left (\frac {1}{2} (c+d x)\right )}{16 \sqrt {1-\cos (c+d x)}}+1024 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _4F_3\left (\frac {3}{2},2,2,2;1,1,\frac {9}{2};2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right )}{6720 d} \] Input:
Integrate[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2,x]
Output:
((a*(1 + Cos[c + d*x]))^(5/2)*((-35*(2080 + 3131*Cos[c + d*x] + 728*Cos[2* (c + d*x)] + 61*Cos[3*(c + d*x)])*Csc[(c + d*x)/2]^6)/8 + (105*ArcTanh[Sqr t[1 - Cos[c + d*x]]]*(1767 + 1252*Cos[c + d*x] + 872*Cos[2*(c + d*x)] + 10 8*Cos[3*(c + d*x)] + Cos[4*(c + d*x)])*Csc[(c + d*x)/2]^6)/(16*Sqrt[1 - Co s[c + d*x]]) + 1024*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{3/2, 2, 2, 2}, { 1, 1, 9/2}, 2*Sin[(c + d*x)/2]^2])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^3)/ (6720*d)
Time = 0.55 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3241, 27, 3042, 3460, 3042, 3252, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3241 |
| \(\displaystyle \frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}-a \int -\frac {1}{2} \sqrt {\cos (c+d x) a+a} (\cos (c+d x) a+5 a) \sec (c+d x)dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} a \int \sqrt {\cos (c+d x) a+a} (\cos (c+d x) a+5 a) \sec (c+d x)dx+\frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+5 a\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {1}{2} a \left (5 a \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx+\frac {2 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (5 a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {1}{2} a \left (\frac {2 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {10 a^2 \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {a^2 \tan (c+d x) \sqrt {a \cos (c+d x)+a}}{d}+\frac {1}{2} a \left (\frac {10 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {2 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )\) |
Input:
Int[(a + a*Cos[c + d*x])^(5/2)*Sec[c + d*x]^2,x]
Output:
(a*((10*a^(3/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/ d + (2*a^2*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/2 + (a^2*Sqrt[a + a*Cos[c + d*x]]*Tan[c + d*x])/d
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b *Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(431\) vs. \(2(82)=164\).
Time = 6.73 (sec) , antiderivative size = 432, normalized size of antiderivative = 4.70
| method | result | size |
| default | \(\frac {a^{\frac {3}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-8 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -10 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +6 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+5 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) a +5 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) a \right )}{\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\right ) \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\) | \(432\) |
Input:
int((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*2^(1/2)*(a*s in(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2-10*ln(4/(2*cos(1/2 *d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+ 1/2*c)^2)^(1/2)*a^(1/2)+2*a))*sin(1/2*d*x+1/2*c)^2*a-10*ln(-4/(2*cos(1/2*d *x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/ 2*c)^2)^(1/2)*a^(1/2)-2*a))*sin(1/2*d*x+1/2*c)^2*a+6*2^(1/2)*(a*sin(1/2*d* x+1/2*c)^2)^(1/2)*a^(1/2)+5*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2) *cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*a +5*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2)*cos(1/2*d*x+1/2*c)-2^(1 /2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*a)/(2*cos(1/2*d*x+1/2*c)- 2^(1/2))/(2*cos(1/2*d*x+1/2*c)+2^(1/2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+ 1/2*c)^2)^(1/2)/d
Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.78 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=\frac {5 \, {\left (a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{4 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^2,x, algorithm="fricas")
Output:
1/4*(5*(a^2*cos(d*x + c)^2 + a^2*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c) ^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(2*a^2*co s(d*x + c) + a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+a*cos(d*x+c))**(5/2)*sec(d*x+c)**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 10847 vs. \(2 (82) = 164\).
Time = 0.42 (sec) , antiderivative size = 10847, normalized size of antiderivative = 117.90 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=\text {Too large to display} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^2,x, algorithm="maxima")
Output:
-1/252*(1449*sqrt(2)*a^2*cos(5/2*d*x + 5/2*c)^3*sin(2*d*x + 2*c) - 63*(sqr t(2)*a^2*cos(2*d*x + 2*c)^2 + sqrt(2)*a^2*sin(2*d*x + 2*c)^2 + 25*sqrt(2)* a^2*cos(2*d*x + 2*c) + 24*sqrt(2)*a^2)*sin(5/2*d*x + 5/2*c)^3 - 252*sqrt(2 )*a^2*sin(1/2*d*x + 1/2*c) + 21*(5*(5*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 1 2*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin( 1/2*d*x + 1/2*c) + 2) + 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2 *c) + 2) - 3*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3* a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*co s(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c) ^2 - 15*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sq rt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 15*a^2* log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/ 2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 15*a^2*log(2*cos(1/ 2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2* c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 15*a^2*log(2*cos(1/2*d*x + 1/2* c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt( 2)*sin(1/2*d*x + 1/2*c) + 2) + 5*(5*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + ...
Time = 0.41 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.47 \[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=-\frac {\sqrt {2} {\left (5 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {4 \, a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{4 \, d} \] Input:
integrate((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^2,x, algorithm="giac")
Output:
-1/4*sqrt(2)*(5*sqrt(2)*a^2*log(abs(-2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c))/a bs(2*sqrt(2) + 4*sin(1/2*d*x + 1/2*c)))*sgn(cos(1/2*d*x + 1/2*c)) - 8*a^2* sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c) + 4*a^2*sgn(cos(1/2*d*x + 1 /2*c))*sin(1/2*d*x + 1/2*c)/(2*sin(1/2*d*x + 1/2*c)^2 - 1))*sqrt(a)/d
Timed out. \[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=\int \frac {{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \] Input:
int((a + a*cos(c + d*x))^(5/2)/cos(c + d*x)^2,x)
Output:
int((a + a*cos(c + d*x))^(5/2)/cos(c + d*x)^2, x)
\[ \int (a+a \cos (c+d x))^{5/2} \sec ^2(c+d x) \, dx=\sqrt {a}\, a^{2} \left (2 \left (\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}d x \right )+\int \sqrt {\cos \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x +\int \sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}d x \right ) \] Input:
int((a+a*cos(d*x+c))^(5/2)*sec(d*x+c)^2,x)
Output:
sqrt(a)*a**2*(2*int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)*sec(c + d*x)**2,x) + int(sqrt(cos(c + d*x) + 1)*cos(c + d*x)**2*sec(c + d*x)**2,x) + int(sqr t(cos(c + d*x) + 1)*sec(c + d*x)**2,x))