Integrand size = 16, antiderivative size = 118 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {\tan (c+d x)}{7 d \left (a \sec ^2(c+d x)\right )^{7/2}}+\frac {6 \tan (c+d x)}{35 a d \left (a \sec ^2(c+d x)\right )^{5/2}}+\frac {8 \tan (c+d x)}{35 a^2 d \left (a \sec ^2(c+d x)\right )^{3/2}}+\frac {16 \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \]
1/7*tan(d*x+c)/d/(a*sec(d*x+c)^2)^(7/2)+6/35*tan(d*x+c)/a/d/(a*sec(d*x+c)^ 2)^(5/2)+8/35*tan(d*x+c)/a^2/d/(a*sec(d*x+c)^2)^(3/2)+16/35*tan(d*x+c)/a^3 /d/(a*sec(d*x+c)^2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {\left (35-35 \sin ^2(c+d x)+21 \sin ^4(c+d x)-5 \sin ^6(c+d x)\right ) \tan (c+d x)}{35 a^3 d \sqrt {a \sec ^2(c+d x)}} \]
((35 - 35*Sin[c + d*x]^2 + 21*Sin[c + d*x]^4 - 5*Sin[c + d*x]^6)*Tan[c + d *x])/(35*a^3*d*Sqrt[a*Sec[c + d*x]^2])
Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4140, 3042, 4610, 209, 209, 209, 208}
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.
| \(\displaystyle \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \tan (c+d x)^2+a\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4140 |
| \(\displaystyle \int \frac {1}{\left (a \sec ^2(c+d x)\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \sec (c+d x)^2\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4610 |
| \(\displaystyle \frac {a \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{9/2}}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {a \left (\frac {6 \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{7/2}}d\tan (c+d x)}{7 a}+\frac {\tan (c+d x)}{7 a \left (a \tan ^2(c+d x)+a\right )^{7/2}}\right )}{d}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {a \left (\frac {6 \left (\frac {4 \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{5/2}}d\tan (c+d x)}{5 a}+\frac {\tan (c+d x)}{5 a \left (a \tan ^2(c+d x)+a\right )^{5/2}}\right )}{7 a}+\frac {\tan (c+d x)}{7 a \left (a \tan ^2(c+d x)+a\right )^{7/2}}\right )}{d}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {a \left (\frac {6 \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^{3/2}}d\tan (c+d x)}{3 a}+\frac {\tan (c+d x)}{3 a \left (a \tan ^2(c+d x)+a\right )^{3/2}}\right )}{5 a}+\frac {\tan (c+d x)}{5 a \left (a \tan ^2(c+d x)+a\right )^{5/2}}\right )}{7 a}+\frac {\tan (c+d x)}{7 a \left (a \tan ^2(c+d x)+a\right )^{7/2}}\right )}{d}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {a \left (\frac {6 \left (\frac {4 \left (\frac {2 \tan (c+d x)}{3 a^2 \sqrt {a \tan ^2(c+d x)+a}}+\frac {\tan (c+d x)}{3 a \left (a \tan ^2(c+d x)+a\right )^{3/2}}\right )}{5 a}+\frac {\tan (c+d x)}{5 a \left (a \tan ^2(c+d x)+a\right )^{5/2}}\right )}{7 a}+\frac {\tan (c+d x)}{7 a \left (a \tan ^2(c+d x)+a\right )^{7/2}}\right )}{d}\) |
(a*(Tan[c + d*x]/(7*a*(a + a*Tan[c + d*x]^2)^(7/2)) + (6*(Tan[c + d*x]/(5* a*(a + a*Tan[c + d*x]^2)^(5/2)) + (4*(Tan[c + d*x]/(3*a*(a + a*Tan[c + d*x ]^2)^(3/2)) + (2*Tan[c + d*x])/(3*a^2*Sqrt[a + a*Tan[c + d*x]^2])))/(5*a)) )/(7*a)))/d
3.3.86.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*sec[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a, b]
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 = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\tan \left (d x +c \right )}{7 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {7}{2}}}+\frac {\frac {6 \tan \left (d x +c \right )}{35 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}\right )}{7 a}}{a}\right )}{d}\) | \(119\) |
| default | \(\frac {a \left (\frac {\tan \left (d x +c \right )}{7 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {7}{2}}}+\frac {\frac {6 \tan \left (d x +c \right )}{35 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 \tan \left (d x +c \right )}{15 a \left (a +a \tan \left (d x +c \right )^{2}\right )^{\frac {3}{2}}}+\frac {8 \tan \left (d x +c \right )}{15 a^{2} \sqrt {a +a \tan \left (d x +c \right )^{2}}}\right )}{7 a}}{a}\right )}{d}\) | \(119\) |
| risch | \(-\frac {i {\mathrm e}^{8 i \left (d x +c \right )}}{896 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}-\frac {35 i {\mathrm e}^{2 i \left (d x +c \right )}}{128 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}+\frac {35 i}{128 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {7 i {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{3}}-\frac {11 i \cos \left (6 d x +6 c \right )}{1120 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {27 \sin \left (6 d x +6 c \right )}{2240 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}-\frac {7 i \cos \left (4 d x +4 c \right )}{160 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}+\frac {21 \sin \left (4 d x +4 c \right )}{320 a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}}\, d}\) | \(447\) |
1/d*a*(1/7/a*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(7/2)+6/7/a*(1/5/a*tan(d*x+c)/( a+a*tan(d*x+c)^2)^(5/2)+4/5/a*(1/3/a*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(3/2)+2 /3/a^2*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(1/2))))
Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {{\left (16 \, \tan \left (d x + c\right )^{7} + 56 \, \tan \left (d x + c\right )^{5} + 70 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} \sqrt {a \tan \left (d x + c\right )^{2} + a}}{35 \, {\left (a^{4} d \tan \left (d x + c\right )^{8} + 4 \, a^{4} d \tan \left (d x + c\right )^{6} + 6 \, a^{4} d \tan \left (d x + c\right )^{4} + 4 \, a^{4} d \tan \left (d x + c\right )^{2} + a^{4} d\right )}} \]
1/35*(16*tan(d*x + c)^7 + 56*tan(d*x + c)^5 + 70*tan(d*x + c)^3 + 35*tan(d *x + c))*sqrt(a*tan(d*x + c)^2 + a)/(a^4*d*tan(d*x + c)^8 + 4*a^4*d*tan(d* x + c)^6 + 6*a^4*d*tan(d*x + c)^4 + 4*a^4*d*tan(d*x + c)^2 + a^4*d)
\[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\int \frac {1}{\left (a \tan ^{2}{\left (c + d x \right )} + a\right )^{\frac {7}{2}}}\, dx \]
Time = 0.40 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {5 \, \sin \left (7 \, d x + 7 \, c\right ) + 49 \, \sin \left (5 \, d x + 5 \, c\right ) + 245 \, \sin \left (3 \, d x + 3 \, c\right ) + 1225 \, \sin \left (d x + c\right )}{2240 \, a^{\frac {7}{2}} d} \]
1/2240*(5*sin(7*d*x + 7*c) + 49*sin(5*d*x + 5*c) + 245*sin(3*d*x + 3*c) + 1225*sin(d*x + c))/(a^(7/2)*d)
Leaf count of result is larger than twice the leaf count of optimal. 2158 vs. \(2 (102) = 204\).
Time = 2.85 (sec) , antiderivative size = 2158, normalized size of antiderivative = 18.29 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\text {Too large to display} \]
-2/35*(35*tan(1/2*d*x)^13*tan(1/2*c)^14 - 245*tan(1/2*d*x)^13*tan(1/2*c)^1 2 - 490*tan(1/2*d*x)^12*tan(1/2*c)^13 + 70*tan(1/2*d*x)^11*tan(1/2*c)^14 + 735*tan(1/2*d*x)^13*tan(1/2*c)^10 + 2940*tan(1/2*d*x)^12*tan(1/2*c)^11 + 3430*tan(1/2*d*x)^11*tan(1/2*c)^12 + 301*tan(1/2*d*x)^9*tan(1/2*c)^14 - 12 25*tan(1/2*d*x)^13*tan(1/2*c)^8 - 7350*tan(1/2*d*x)^12*tan(1/2*c)^9 - 1813 0*tan(1/2*d*x)^11*tan(1/2*c)^10 - 19600*tan(1/2*d*x)^10*tan(1/2*c)^11 - 52 43*tan(1/2*d*x)^9*tan(1/2*c)^12 - 2450*tan(1/2*d*x)^8*tan(1/2*c)^13 + 212* tan(1/2*d*x)^7*tan(1/2*c)^14 + 1225*tan(1/2*d*x)^13*tan(1/2*c)^6 + 9800*ta n(1/2*d*x)^12*tan(1/2*c)^7 + 36750*tan(1/2*d*x)^11*tan(1/2*c)^8 + 78400*ta n(1/2*d*x)^10*tan(1/2*c)^9 + 84721*tan(1/2*d*x)^9*tan(1/2*c)^10 + 34300*ta n(1/2*d*x)^8*tan(1/2*c)^11 + 11284*tan(1/2*d*x)^7*tan(1/2*c)^12 + 301*tan( 1/2*d*x)^5*tan(1/2*c)^14 - 735*tan(1/2*d*x)^13*tan(1/2*c)^4 - 7350*tan(1/2 *d*x)^12*tan(1/2*c)^5 - 36750*tan(1/2*d*x)^11*tan(1/2*c)^6 - 117600*tan(1/ 2*d*x)^10*tan(1/2*c)^7 - 230055*tan(1/2*d*x)^9*tan(1/2*c)^8 - 240590*tan(1 /2*d*x)^8*tan(1/2*c)^9 - 113148*tan(1/2*d*x)^7*tan(1/2*c)^10 - 39200*tan(1 /2*d*x)^6*tan(1/2*c)^11 - 5243*tan(1/2*d*x)^5*tan(1/2*c)^12 - 1470*tan(1/2 *d*x)^4*tan(1/2*c)^13 + 70*tan(1/2*d*x)^3*tan(1/2*c)^14 + 245*tan(1/2*d*x) ^13*tan(1/2*c)^2 + 2940*tan(1/2*d*x)^12*tan(1/2*c)^3 + 18130*tan(1/2*d*x)^ 11*tan(1/2*c)^4 + 78400*tan(1/2*d*x)^10*tan(1/2*c)^5 + 230055*tan(1/2*d*x) ^9*tan(1/2*c)^6 + 417480*tan(1/2*d*x)^8*tan(1/2*c)^7 + 424900*tan(1/2*d...
Time = 11.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^{7/2}} \, dx=\frac {16\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}+\frac {8\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^2}+\frac {6\,\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{35\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^3}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a}}{7\,a^4\,d\,{\left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}^4} \]
(16*tan(c + d*x)*(a + a*tan(c + d*x)^2)^(1/2))/(35*a^4*d*(tan(c + d*x)^2 + 1)) + (8*tan(c + d*x)*(a + a*tan(c + d*x)^2)^(1/2))/(35*a^4*d*(tan(c + d* x)^2 + 1)^2) + (6*tan(c + d*x)*(a + a*tan(c + d*x)^2)^(1/2))/(35*a^4*d*(ta n(c + d*x)^2 + 1)^3) + (tan(c + d*x)*(a + a*tan(c + d*x)^2)^(1/2))/(7*a^4* d*(tan(c + d*x)^2 + 1)^4)