Integrand size = 25, antiderivative size = 137 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]
1/4*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)+3/16*sec(d*x+c)^( 3/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+3/32*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))/a^(5/2)/d*2^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(308\) vs. \(2(137)=274\).
Time = 0.71 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.25 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {6 \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+14 \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-3 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-6 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-3 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)+6 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)+6 \arcsin \left (\sqrt {\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{32 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]
(6*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 14*Sqrt[1 - Se c[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] - 3*Sqrt[2]*ArcTan[(Sqrt[2]*Sq rt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 6*Sqrt[2]*ArcTan[ (Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] - 3*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] + 6*ArcSin[Sqrt[1 - Sec[c + d*x]]]*(1 + Sec [c + d*x])^2*Tan[c + d*x] + 6*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 = 0.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4297, 3042, 4297, 3042, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4297 |
| \(\displaystyle \frac {3 \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4297 |
| \(\displaystyle \frac {3 \left (\frac {\int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 \left (\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle \frac {3 \left (\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {\int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{2 a d}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}+\frac {\sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
(Sec[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + (3*(A rcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]/(2*Sqrt[2]*a^(3/2)*d) + (Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d* (a + a*Sec[c + d*x])^(3/2))))/(8*a)
3.3.60.3.1 Defintions of rubi rules used
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[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[b*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[d*((m + 1)/(b*(2*m + 1))) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ [{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + n, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(303\) vs. \(2(112)=224\).
Time = 1.26 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.22
| method | result | size |
| default | \(\frac {\left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {5}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3} \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (-2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-5 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{32 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )^{2} \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\) | \(304\) |
1/32/d/a^3*(-((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c )^2-1))^(5/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^3*(-2*a/((1-cos(d*x+c))^2* csc(d*x+c)^2-1))^(1/2)*(-2*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-cos (d*x+c))^3*csc(d*x+c)^3-5*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d *x+c)+csc(d*x+c))+3*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-co t(d*x+c)+csc(d*x+c))))/((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^2/(-(1-cos(d*x+c) )^2*csc(d*x+c)^2-1)^(1/2)
Time = 0.32 (sec) , antiderivative size = 426, normalized size of antiderivative = 3.11 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 7 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 7 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]
[1/64*(3*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)) + 4*(3*cos(d*x + c)^2 + 7*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*(3*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))) - 2*(3*cos(d*x + c)^2 + 7*c os(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. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 84332 vs. \(2 (112) = 224\).
Time = 14.47 (sec) , antiderivative size = 84332, normalized size of antiderivative = 615.56 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
1/32*(512*((2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos( 5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 2*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c ) + (2*cos(2*d*x + 2*c) + cos(d*x + c))*sin(5/2*d*x + 5/2*c) + cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) + 2*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos( 5*d*x + 5*c)^2 + 2560*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)*s in(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)*sin( 5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(8/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2 *d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d *x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2* c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2 *d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)* sin(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3* d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(6/5*arctan2(sin(5/2*d*x + 5/2*c), cos (5/2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5 /2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*c os(2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x ...
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]