Integrand size = 30, antiderivative size = 96 \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx=-\frac {4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \log (\cos (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
-4*c^3*tan(f*x+e)/f/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(1/2)+c^3*ln(c os(f*x+e))*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
Time = 0.56 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68 \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx=\frac {c^3 (-4+\log (\cos (e+f x))+\log (\cos (e+f x)) \sec (e+f x)) \tan (e+f x)}{f (a (1+\sec (e+f x)))^{3/2} \sqrt {c-c \sec (e+f x)}} \]
(c^3*(-4 + Log[Cos[e + f*x]] + Log[Cos[e + f*x]]*Sec[e + f*x])*Tan[e + f*x ])/(f*(a*(1 + Sec[e + f*x]))^(3/2)*Sqrt[c - c*Sec[e + f*x]])
Time = 0.54 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 4398, 3042, 4393, 25, 3042, 3956}
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.
\(\displaystyle \int \frac {(c-c \sec (e+f x))^{5/2}}{(a \sec (e+f x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4398 |
\(\displaystyle \frac {c^2 \int \sqrt {\sec (e+f x) a+a} \sqrt {c-c \sec (e+f x)}dx}{a^2}-\frac {4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {c^2 \int \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a} \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 4393 |
\(\displaystyle \frac {c^3 \tan (e+f x) \int -\tan (e+f x)dx}{a \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {c^3 \tan (e+f x) \int \tan (e+f x)dx}{a \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {c^3 \tan (e+f x) \int \tan (e+f x)dx}{a \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {c^3 \tan (e+f x) \log (\cos (e+f x))}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}\) |
(-4*c^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x ]]) + (c^3*Log[Cos[e + f*x]]*Tan[e + f*x])/(a*f*Sqrt[a + a*Sec[e + f*x]]*S qrt[c - c*Sec[e + f*x]])
3.2.18.3.1 Defintions of rubi rules used
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_))^(m_), x_Symbol] :> Simp[((-a)*c)^(m + 1/2)*(Cot[e + f*x]/(Sqrt[ a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Int[Cot[e + f*x]^(2*m), x] , x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b ^2, 0] && IntegerQ[m + 1/2]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(5/2)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[-8*a^3*Cot[e + f*x]*((c + d*Csc[e + f *x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a^2/c^2 Int[Sqr t[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1) ]
Leaf count of result is larger than twice the leaf count of optimal. \(218\) vs. \(2(88)=176\).
Time = 1.96 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.28
method | result | size |
default | \(\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{3} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}} \sin \left (f x +e \right )^{5} \left (2 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-\ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right )+\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )\right )}{2 f \,a^{2} \left (1-\cos \left (f x +e \right )\right )^{5}}\) | \(219\) |
risch | \(-\frac {c^{2} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+{\mathrm e}^{2 i \left (f x +e \right )} f x +2 \,{\mathrm e}^{2 i \left (f x +e \right )} e +2 i {\mathrm e}^{i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+2 \,{\mathrm e}^{i \left (f x +e \right )} f x +8 i {\mathrm e}^{i \left (f x +e \right )}+4 \,{\mathrm e}^{i \left (f x +e \right )} e +i \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )+f x +2 e \right )}{a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\) | \(229\) |
1/2/f*2^(1/2)/a^2*(-2*a/((1-cos(f*x+e))^2*csc(f*x+e)^2-1))^(1/2)*((1-cos(f *x+e))^2*csc(f*x+e)^2-1)^3*(c*(1-cos(f*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+e )^2-1)*csc(f*x+e)^2)^(5/2)/(1-cos(f*x+e))^5*sin(f*x+e)^5*(2*(1-cos(f*x+e)) ^2*csc(f*x+e)^2+ln(-cot(f*x+e)+csc(f*x+e)+1)-ln((1-cos(f*x+e))^2*csc(f*x+e )^2+1)+ln(-cot(f*x+e)+csc(f*x+e)-1))
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (88) = 176\).
Time = 0.34 (sec) , antiderivative size = 453, normalized size of antiderivative = 4.72 \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx=\left [-\frac {4 \, c^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c \cos \left (f x + e\right )^{4} - {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {-\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + c}{2 \, \cos \left (f x + e\right )^{2}}\right )}{2 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}}, -\frac {2 \, c^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \sqrt {\frac {c}{a}} \arctan \left (\frac {\sqrt {\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{c \cos \left (f x + e\right )^{2} + c}\right )}{a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f}\right ] \]
[-1/2*(4*c^2*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - (a*c^2*cos(f*x + e)^2 + 2*a *c^2*cos(f*x + e) + a*c^2)*sqrt(-c/a)*log(1/2*(c*cos(f*x + e)^4 - (cos(f*x + e)^3 + cos(f*x + e))*sqrt(-c/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)) *sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*sin(f*x + e) + c)/cos(f*x + e)^2) )/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f), -(2*c^2*sqrt((a*c os(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos (f*x + e)*sin(f*x + e) - (a*c^2*cos(f*x + e)^2 + 2*a*c^2*cos(f*x + e) + a* c^2)*sqrt(c/a)*arctan(sqrt(c/a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sq rt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(c*cos(f*x + e)^2 + c)))/(a^2*f*cos(f*x + e)^2 + 2*a^2*f*cos(f*x + e) + a^2*f)]
Timed out. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Timed out. \[ \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]