Integrand size = 25, antiderivative size = 174 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {43 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {11 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \] Output:
2*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-43/32*arcta nh(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)) *2^(1/2)/a^(5/2)/d-1/4*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2 )-11/16*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)
Time = 0.47 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.77 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {-22 \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-30 \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+43 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)+86 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)+43 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)-22 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)-86 \arcsin \left (\sqrt {\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{32 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \] Input:
Integrate[Sec[c + d*x]^(7/2)/(a + a*Sec[c + d*x])^(5/2),x]
Output:
(-22*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] - 30*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] + 43*Sqrt[2]*ArcTan[(Sqrt[2] *Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + 86*Sqrt[2]*Arc Tan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[ c + d*x] + 43*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^2*Tan[c + d*x] - 22*ArcSin[Sqrt[1 - Sec[c + d*x]]]*(1 + Sec[c + d*x])^2*Tan[c + d*x] - 86*ArcSin[Sqrt[Sec[c + d*x]]]*(1 + Sec[c + d*x])^2*Tan[c + d*x])/(32*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x] ))^(5/2))
Time = 1.05 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4303, 27, 3042, 4507, 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.
\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4303 |
\(\displaystyle -\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (3 a-8 a \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (3 a-8 a \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 a-8 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle -\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left (11 a^2-32 a^2 \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left (11 a^2-32 a^2 \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (11 a^2-32 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4511 |
\(\displaystyle -\frac {\frac {43 a^2 \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-32 a \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {43 a^2 \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-32 a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4288 |
\(\displaystyle -\frac {\frac {43 a^2 \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {64 a \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 222 |
\(\displaystyle -\frac {\frac {43 a^2 \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {64 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4295 |
\(\displaystyle -\frac {\frac {-\frac {86 a^2 \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {64 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\frac {43 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {64 a^{3/2} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {11 a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
Input:
Int[Sec[c + d*x]^(7/2)/(a + a*Sec[c + d*x])^(5/2),x]
Output:
-1/4*(Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) - (( (-64*a^(3/2)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (43*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sq rt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/(4*a^2) + (11*a*Sec[c + d*x]^(3/2)*Si n[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)))/(8*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_), x_Symbol] :> Simp[(-d^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d *Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Simp[d^2/(a*b*(2*m + 1)) Int[ (a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(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 - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[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[(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]
Time = 3.00 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.61
method | result | size |
default | \(-\frac {\sec \left (d x +c \right )^{\frac {7}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{4} \sqrt {-a \left (-1-\sec \left (d x +c \right )\right )}\, \left (2 \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-16 \sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-16 \sqrt {2}\, \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+13 \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+43 \arctan \left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \sqrt {2}}{64 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )^{3} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(280\) |
Input:
int(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/64/d/a^3*sec(d*x+c)^(7/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^4*(-a*(-1-s ec(d*x+c)))^(1/2)*(2*(-2/(cos(d*x+c)+1))^(1/2)*(1-cos(d*x+c))^3*csc(d*x+c) ^3-16*2^(1/2)*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/ 2))-16*2^(1/2)*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c) +1))+13*(-2/(cos(d*x+c)+1))^(1/2)*(-cot(d*x+c)+csc(d*x+c))+43*arctan(1/2*2 ^(1/2)*(cot(d*x+c)-csc(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2)))/((1-cos(d*x+c)) ^2*csc(d*x+c)^2+1)^3*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (143) = 286\).
Time = 0.12 (sec) , antiderivative size = 662, normalized size of antiderivative = 3.80 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[1/64*(43*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1) *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)) + 32*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*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))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 4*(11*cos(d*x + c)^2 + 15*cos(d*x + c))*sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(43* sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(-a)* arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d *x + c))/(a*sin(d*x + c))) + 32*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos (d*x + c) + 1)*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*(11*cos(d*x + c)^2 + 15*cos(d*x + c))*sqrt((a*cos(d*x + c) + a) /cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3* a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(7/2)/(a+a*sec(d*x+c))**(5/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 4988 vs. \(2 (143) = 286\).
Time = 0.44 (sec) , antiderivative size = 4988, normalized size of antiderivative = 28.67 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
1/32*(44*(sin(4*d*x + 4*c) + 6*sin(2*d*x + 2*c) + 4*sin(3/2*arctan2(sin(2* d*x + 2*c), cos(2*d*x + 2*c))) + 4*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c))))*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 16*( 19*sin(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 19*sin(3/4*arcta n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 11*sin(1/4*arctan2(sin(2*d*x + 2 *c), cos(2*d*x + 2*c))))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c ))) + 76*(sin(4*d*x + 4*c) + 6*sin(2*d*x + 2*c) + 4*sin(1/2*arctan2(sin(2* d*x + 2*c), cos(2*d*x + 2*c))))*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d* x + 2*c))) - 76*(sin(4*d*x + 4*c) + 6*sin(2*d*x + 2*c) + 4*sin(1/2*arctan2 (sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(3/4*arctan2(sin(2*d*x + 2*c), c os(2*d*x + 2*c))) - 44*(sin(4*d*x + 4*c) + 6*sin(2*d*x + 2*c))*cos(1/4*arc tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 16*(sqrt(2)*cos(4*d*x + 4*c)^2 + 36*sqrt(2)*cos(2*d*x + 2*c)^2 + 16*sqrt(2)*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 16*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sqrt(2)*sin(4*d*x + 4*c)^2 + 12*sqrt(2)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 36*sqrt(2)*sin(2*d*x + 2*c)^2 + 16*sqrt(2)*sin(3/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 16*sqrt(2)*sin(1/2*arct an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*(6*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*cos(4*d*x + 4*c) + 8*(sqrt(2)*cos(4*d*x + 4*c) + 6*sqrt(2)*cos (2*d*x + 2*c) + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +...
Time = 1.68 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {2 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} + \frac {13 \, \sqrt {2}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {43 \, \sqrt {2} \log \left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a^{\frac {5}{2}}} - \frac {64 \, \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{a^{\frac {5}{2}}} + \frac {64 \, \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{a^{\frac {5}{2}}}}{64 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:
integrate(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
-1/64*(2*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c )^2/a^3 + 13*sqrt(2)/a^3)*tan(1/2*d*x + 1/2*c) - 43*sqrt(2)*log((sqrt(a)*t an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/a^(5/2) - 64* 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^(5/2) + 64*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^(5/2))/ (d*sgn(cos(d*x + c)))
Timed out. \[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((1/cos(c + d*x))^(7/2)/(a + a/cos(c + d*x))^(5/2),x)
Output:
int((1/cos(c + d*x))^(7/2)/(a + a/cos(c + d*x))^(5/2), x)
\[ \int \frac {\sec ^{\frac {7}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int(sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3)/( sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x))/a**3