Integrand size = 21, antiderivative size = 67 \[ \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 (d \tan (a+b x))^{5/2}}{5 b d}+\frac {4 (d \tan (a+b x))^{9/2}}{9 b d^3}+\frac {2 (d \tan (a+b x))^{13/2}}{13 b d^5} \]
2/5*(d*tan(b*x+a))^(5/2)/b/d+4/9*(d*tan(b*x+a))^(9/2)/b/d^3+2/13*(d*tan(b* x+a))^(13/2)/b/d^5
Time = 0.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78 \[ \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 d \left (-32-8 \sec ^2(a+b x)-5 \sec ^4(a+b x)+45 \sec ^6(a+b x)\right ) \sqrt {d \tan (a+b x)}}{585 b} \]
(2*d*(-32 - 8*Sec[a + b*x]^2 - 5*Sec[a + b*x]^4 + 45*Sec[a + b*x]^6)*Sqrt[ d*Tan[a + b*x]])/(585*b)
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3087, 244, 2009}
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.
\(\displaystyle \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (a+b x)^6 (d \tan (a+b x))^{3/2}dx\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\int (d \tan (a+b x))^{3/2} \left (\tan ^2(a+b x)+1\right )^2d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\frac {(d \tan (a+b x))^{11/2}}{d^4}+\frac {2 (d \tan (a+b x))^{7/2}}{d^2}+(d \tan (a+b x))^{3/2}\right )d\tan (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 (d \tan (a+b x))^{13/2}}{13 d^5}+\frac {4 (d \tan (a+b x))^{9/2}}{9 d^3}+\frac {2 (d \tan (a+b x))^{5/2}}{5 d}}{b}\) |
((2*(d*Tan[a + b*x])^(5/2))/(5*d) + (4*(d*Tan[a + b*x])^(9/2))/(9*d^3) + ( 2*(d*Tan[a + b*x])^(13/2))/(13*d^5))/b
3.3.36.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Time = 0.39 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (d \tan \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {4 d^{2} \left (d \tan \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 d^{4} \left (d \tan \left (b x +a \right )\right )^{\frac {5}{2}}}{5}}{d^{5} b}\) | \(52\) |
default | \(\frac {\frac {2 \left (d \tan \left (b x +a \right )\right )^{\frac {13}{2}}}{13}+\frac {4 d^{2} \left (d \tan \left (b x +a \right )\right )^{\frac {9}{2}}}{9}+\frac {2 d^{4} \left (d \tan \left (b x +a \right )\right )^{\frac {5}{2}}}{5}}{d^{5} b}\) | \(52\) |
2/d^5/b*(1/13*(d*tan(b*x+a))^(13/2)+2/9*d^2*(d*tan(b*x+a))^(9/2)+1/5*d^4*( d*tan(b*x+a))^(5/2))
Time = 0.30 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {2 \, {\left (32 \, d \cos \left (b x + a\right )^{6} + 8 \, d \cos \left (b x + a\right )^{4} + 5 \, d \cos \left (b x + a\right )^{2} - 45 \, d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{585 \, b \cos \left (b x + a\right )^{6}} \]
-2/585*(32*d*cos(b*x + a)^6 + 8*d*cos(b*x + a)^4 + 5*d*cos(b*x + a)^2 - 45 *d)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*cos(b*x + a)^6)
\[ \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sec ^{6}{\left (a + b x \right )}\, dx \]
Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 \, {\left (45 \, \left (d \tan \left (b x + a\right )\right )^{\frac {13}{2}} + 130 \, \left (d \tan \left (b x + a\right )\right )^{\frac {9}{2}} d^{2} + 117 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{4}\right )}}{585 \, b d^{5}} \]
2/585*(45*(d*tan(b*x + a))^(13/2) + 130*(d*tan(b*x + a))^(9/2)*d^2 + 117*( d*tan(b*x + a))^(5/2)*d^4)/(b*d^5)
Time = 0.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16 \[ \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {2 \, {\left (45 \, \sqrt {d \tan \left (b x + a\right )} d^{6} \tan \left (b x + a\right )^{6} + 130 \, \sqrt {d \tan \left (b x + a\right )} d^{6} \tan \left (b x + a\right )^{4} + 117 \, \sqrt {d \tan \left (b x + a\right )} d^{6} \tan \left (b x + a\right )^{2}\right )}}{585 \, b d^{5}} \]
2/585*(45*sqrt(d*tan(b*x + a))*d^6*tan(b*x + a)^6 + 130*sqrt(d*tan(b*x + a ))*d^6*tan(b*x + a)^4 + 117*sqrt(d*tan(b*x + a))*d^6*tan(b*x + a)^2)/(b*d^ 5)
Time = 11.54 (sec) , antiderivative size = 392, normalized size of antiderivative = 5.85 \[ \int \sec ^6(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {64\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{585\,b}-\frac {64\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{585\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}-\frac {32\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{195\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {1216\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{117\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3}-\frac {3488\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{117\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {384\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{13\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {128\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{13\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^6} \]
(1216*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2) )/(117*b*(exp(a*2i + b*x*2i) + 1)^3) - (64*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(585*b*(exp(a*2i + b*x*2i) + 1)) - (32*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/ (195*b*(exp(a*2i + b*x*2i) + 1)^2) - (64*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1 i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(585*b) - (3488*d*(-(d*(exp(a*2i + b* x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(117*b*(exp(a*2i + b*x*2i ) + 1)^4) + (384*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(13*b*(exp(a*2i + b*x*2i) + 1)^5) - (128*d*(-(d*(exp(a*2i + b *x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(13*b*(exp(a*2i + b*x*2i ) + 1)^6)