Integrand size = 10, antiderivative size = 65 \[ \int \left (a \sec ^2(x)\right )^{5/2} \, dx=\frac {3}{8} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a \sec ^2(x)}}\right )+\frac {3}{8} a^2 \sqrt {a \sec ^2(x)} \tan (x)+\frac {1}{4} a \left (a \sec ^2(x)\right )^{3/2} \tan (x) \]
3/8*a^(5/2)*arctanh(a^(1/2)*tan(x)/(a*sec(x)^2)^(1/2))+1/4*a*(a*sec(x)^2)^ (3/2)*tan(x)+3/8*a^2*(a*sec(x)^2)^(1/2)*tan(x)
Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.57 \[ \int \left (a \sec ^2(x)\right )^{5/2} \, dx=\frac {1}{8} \cos (x) \left (a \sec ^2(x)\right )^{5/2} \left (3 \text {arctanh}(\sin (x)) \cos ^4(x)+\left (2+3 \cos ^2(x)\right ) \sin (x)\right ) \]
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4610, 211, 211, 224, 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.
\(\displaystyle \int \left (a \sec ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \sec (x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 4610 |
\(\displaystyle a \int \left (a \tan ^2(x)+a\right )^{3/2}d\tan (x)\) |
\(\Big \downarrow \) 211 |
\(\displaystyle a \left (\frac {3}{4} a \int \sqrt {a \tan ^2(x)+a}d\tan (x)+\frac {1}{4} \tan (x) \left (a \tan ^2(x)+a\right )^{3/2}\right )\) |
\(\Big \downarrow \) 211 |
\(\displaystyle a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {a \tan ^2(x)+a}}d\tan (x)+\frac {1}{2} \tan (x) \sqrt {a \tan ^2(x)+a}\right )+\frac {1}{4} \tan (x) \left (a \tan ^2(x)+a\right )^{3/2}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {a \tan ^2(x)}{a \tan ^2(x)+a}}d\frac {\tan (x)}{\sqrt {a \tan ^2(x)+a}}+\frac {1}{2} \tan (x) \sqrt {a \tan ^2(x)+a}\right )+\frac {1}{4} \tan (x) \left (a \tan ^2(x)+a\right )^{3/2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a \left (\frac {3}{4} a \left (\frac {1}{2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \tan (x)}{\sqrt {a \tan ^2(x)+a}}\right )+\frac {1}{2} \tan (x) \sqrt {a \tan ^2(x)+a}\right )+\frac {1}{4} \tan (x) \left (a \tan ^2(x)+a\right )^{3/2}\right )\) |
a*((Tan[x]*(a + a*Tan[x]^2)^(3/2))/4 + (3*a*((Sqrt[a]*ArcTanh[(Sqrt[a]*Tan [x])/Sqrt[a + a*Tan[x]^2]])/2 + (Tan[x]*Sqrt[a + a*Tan[x]^2])/2))/4)
3.1.48.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFac tors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] && !IntegerQ[p]
Time = 1.99 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {\sqrt {a \sec \left (x \right )^{2}}\, a^{2} \left (3 \cos \left (x \right ) \ln \left (-\cot \left (x \right )+\csc \left (x \right )+1\right )-3 \cos \left (x \right ) \ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )+3 \tan \left (x \right )+2 \sec \left (x \right )^{2} \tan \left (x \right )\right )}{8}\) | \(53\) |
risch | \(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \left (3 \,{\mathrm e}^{6 i x}+11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}-3\right )}{4 \left ({\mathrm e}^{2 i x}+1\right )^{3}}+\frac {3 a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}+i\right ) \cos \left (x \right )}{4}-\frac {3 a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}}\, \ln \left ({\mathrm e}^{i x}-i\right ) \cos \left (x \right )}{4}\) | \(126\) |
1/8*(a*sec(x)^2)^(1/2)*a^2*(3*cos(x)*ln(-cot(x)+csc(x)+1)-3*cos(x)*ln(-cot (x)+csc(x)-1)+3*tan(x)+2*sec(x)^2*tan(x))
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \left (a \sec ^2(x)\right )^{5/2} \, dx=-\frac {{\left (3 \, a^{2} \cos \left (x\right )^{4} \log \left (-\frac {\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right ) - 2 \, {\left (3 \, a^{2} \cos \left (x\right )^{2} + 2 \, a^{2}\right )} \sin \left (x\right )\right )} \sqrt {\frac {a}{\cos \left (x\right )^{2}}}}{16 \, \cos \left (x\right )^{3}} \]
-1/16*(3*a^2*cos(x)^4*log(-(sin(x) - 1)/(sin(x) + 1)) - 2*(3*a^2*cos(x)^2 + 2*a^2)*sin(x))*sqrt(a/cos(x)^2)/cos(x)^3
\[ \int \left (a \sec ^2(x)\right )^{5/2} \, dx=\int \left (a \sec ^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1111 vs. \(2 (49) = 98\).
Time = 0.53 (sec) , antiderivative size = 1111, normalized size of antiderivative = 17.09 \[ \int \left (a \sec ^2(x)\right )^{5/2} \, dx=\text {Too large to display} \]
1/16*(176*a^2*cos(3*x)*sin(2*x) + 48*a^2*cos(x)*sin(2*x) - 48*a^2*cos(2*x) *sin(x) - 12*a^2*sin(x) + 4*(3*a^2*sin(7*x) + 11*a^2*sin(5*x) - 11*a^2*sin (3*x) - 3*a^2*sin(x))*cos(8*x) - 24*(2*a^2*sin(6*x) + 3*a^2*sin(4*x) + 2*a ^2*sin(2*x))*cos(7*x) + 16*(11*a^2*sin(5*x) - 11*a^2*sin(3*x) - 3*a^2*sin( x))*cos(6*x) - 88*(3*a^2*sin(4*x) + 2*a^2*sin(2*x))*cos(5*x) - 24*(11*a^2* sin(3*x) + 3*a^2*sin(x))*cos(4*x) + 3*(a^2*cos(8*x)^2 + 16*a^2*cos(6*x)^2 + 36*a^2*cos(4*x)^2 + 16*a^2*cos(2*x)^2 + a^2*sin(8*x)^2 + 16*a^2*sin(6*x) ^2 + 36*a^2*sin(4*x)^2 + 48*a^2*sin(4*x)*sin(2*x) + 16*a^2*sin(2*x)^2 + 8* a^2*cos(2*x) + a^2 + 2*(4*a^2*cos(6*x) + 6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(8*x) + 8*(6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(6*x) + 12*( 4*a^2*cos(2*x) + a^2)*cos(4*x) + 4*(2*a^2*sin(6*x) + 3*a^2*sin(4*x) + 2*a^ 2*sin(2*x))*sin(8*x) + 16*(3*a^2*sin(4*x) + 2*a^2*sin(2*x))*sin(6*x))*log( cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) - 3*(a^2*cos(8*x)^2 + 16*a^2*cos(6*x)^ 2 + 36*a^2*cos(4*x)^2 + 16*a^2*cos(2*x)^2 + a^2*sin(8*x)^2 + 16*a^2*sin(6* x)^2 + 36*a^2*sin(4*x)^2 + 48*a^2*sin(4*x)*sin(2*x) + 16*a^2*sin(2*x)^2 + 8*a^2*cos(2*x) + a^2 + 2*(4*a^2*cos(6*x) + 6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(8*x) + 8*(6*a^2*cos(4*x) + 4*a^2*cos(2*x) + a^2)*cos(6*x) + 12 *(4*a^2*cos(2*x) + a^2)*cos(4*x) + 4*(2*a^2*sin(6*x) + 3*a^2*sin(4*x) + 2* a^2*sin(2*x))*sin(8*x) + 16*(3*a^2*sin(4*x) + 2*a^2*sin(2*x))*sin(6*x))*lo g(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1) - 4*(3*a^2*cos(7*x) + 11*a^2*cos(...
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \left (a \sec ^2(x)\right )^{5/2} \, dx=\frac {1}{16} \, {\left (3 \, a^{2} \log \left (\sin \left (x\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (x\right )\right ) - 3 \, a^{2} \log \left (-\sin \left (x\right ) + 1\right ) \mathrm {sgn}\left (\cos \left (x\right )\right ) - \frac {2 \, {\left (3 \, a^{2} \mathrm {sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )^{3} - 5 \, a^{2} \mathrm {sgn}\left (\cos \left (x\right )\right ) \sin \left (x\right )\right )}}{{\left (\sin \left (x\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a} \]
1/16*(3*a^2*log(sin(x) + 1)*sgn(cos(x)) - 3*a^2*log(-sin(x) + 1)*sgn(cos(x )) - 2*(3*a^2*sgn(cos(x))*sin(x)^3 - 5*a^2*sgn(cos(x))*sin(x))/(sin(x)^2 - 1)^2)*sqrt(a)
Timed out. \[ \int \left (a \sec ^2(x)\right )^{5/2} \, dx=\int {\left (\frac {a}{{\cos \left (x\right )}^2}\right )}^{5/2} \,d x \]