Integrand size = 23, antiderivative size = 227 \[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\frac {2 a b (d \cos (e+f x))^m \operatorname {Hypergeometric2F1}\left (2,-\frac {m}{2},1-\frac {m}{2},\frac {b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{\left (a^2+b^2\right )^2 f m}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},2,\frac {2+m}{2},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan (e+f x)}{a^2 f}+\frac {b^2 \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {2+m}{2},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan ^3(e+f x)}{3 a^4 f} \]
2*a*b*(d*cos(f*x+e))^m*hypergeom([2, -1/2*m],[1-1/2*m],b^2*sec(f*x+e)^2/(a ^2+b^2))/(a^2+b^2)^2/f/m+AppellF1(1/2,2,1+1/2*m,3/2,b^2*tan(f*x+e)^2/a^2,- tan(f*x+e)^2)*(d*cos(f*x+e))^m*(sec(f*x+e)^2)^(1/2*m)*tan(f*x+e)/a^2/f+1/3 *b^2*AppellF1(3/2,2,1+1/2*m,5/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*cos (f*x+e))^m*(sec(f*x+e)^2)^(1/2*m)*tan(f*x+e)^3/a^4/f
Result contains complex when optimal does not.
Time = 17.71 (sec) , antiderivative size = 2502, normalized size of antiderivative = 11.02 \[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\text {Result too large to show} \]
((d*Cos[e + f*x])^m*((2*a*b*(-1 + (Sec[e + f*x]^2)^(-1/2*m)))/m + a^2*Hype rgeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x] - b^2*Hyper geometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x] - (2*a*b*App ellF1[m, m/2, m/2, 1 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b *Tan[e + f*x])]*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*((b*( I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))/(m*(Sec[e + f*x]^2)^(m/2)) - (b*(a^2 + b^2)*AppellF1[1 + m, m/2, m/2, 2 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*((b*(-I + Tan[e + f*x]))/(a + b*T an[e + f*x]))^(m/2)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))/( (1 + m)*(Sec[e + f*x]^2)^(m/2)*(a + b*Tan[e + f*x]))))/(f*(a + b*Tan[e + f *x])^2*(a^2*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Sec[e + f*x]^2 - b^2*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Sec[e + f*x]^2 - (2*a*b*Tan[e + f*x])/(Sec[e + f*x]^2)^(m/2) + (2*a*b*AppellF1[m, m/2, m/2, 1 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*Tan[e + f*x]*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)* ((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))/(Sec[e + f*x]^2)^(m/2 ) + (b^2*(a^2 + b^2)*AppellF1[1 + m, m/2, m/2, 2 + m, (a - I*b)/(a + b*Tan [e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*(Sec[e + f*x]^2)^(1 - m/2)*((b *(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*((b*(I + Tan[e + f*x]))/ (a + b*Tan[e + f*x]))^(m/2))/((1 + m)*(a + b*Tan[e + f*x])^2) + (b*(a^2...
Time = 0.58 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3998, 3042, 3994, 505, 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.
\(\displaystyle \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle (d \cos (e+f x))^m (d \sec (e+f x))^m \int \frac {(d \sec (e+f x))^{-m}}{(a+b \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (d \cos (e+f x))^m (d \sec (e+f x))^m \int \frac {(d \sec (e+f x))^{-m}}{(a+b \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 3994 |
\(\displaystyle \frac {\sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \int \frac {\left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{(a+b \tan (e+f x))^2}d(b \tan (e+f x))}{b f}\) |
\(\Big \downarrow \) 505 |
\(\displaystyle \frac {\sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \int \left (\frac {a^2 \left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{\left (a^2-b^2 \tan ^2(e+f x)\right )^2}-\frac {2 a b \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{\left (a^2-b^2 \tan ^2(e+f x)\right )^2}+\frac {b^2 \tan ^2(e+f x) \left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{\left (b^2 \tan ^2(e+f x)-a^2\right )^2}\right )d(b \tan (e+f x))}{b f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},2,\frac {m+2}{2},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2}+\frac {2 a b^2 \left (\tan ^2(e+f x)+1\right )^{-m/2} \operatorname {Hypergeometric2F1}\left (2,-\frac {m}{2},1-\frac {m}{2},\frac {\tan ^2(e+f x) b^2+b^2}{a^2+b^2}\right )}{m \left (a^2+b^2\right )^2}+\frac {b^3 \tan ^3(e+f x) \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {m+2}{2},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4}\right )}{b f}\) |
((d*Cos[e + f*x])^m*(Sec[e + f*x]^2)^(m/2)*((b*AppellF1[1/2, 2, (2 + m)/2, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x])/a^2 + (b^3* AppellF1[3/2, 2, (2 + m)/2, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2 ]*Tan[e + f*x]^3)/(3*a^4) + (2*a*b^2*Hypergeometric2F1[2, -1/2*m, 1 - m/2, (b^2 + b^2*Tan[e + f*x]^2)/(a^2 + b^2)])/((a^2 + b^2)^2*m*(1 + Tan[e + f* x]^2)^(m/2))))/(b*f)
3.7.99.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E xpandIntegrand[(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^( -n), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && PosQ[a/b]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2])) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] && !IntegerQ[m] && IntegerQ[n]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m Int[( a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m , n}, x] && !IntegerQ[m]
\[\int \frac {\left (d \cos \left (f x +e \right )\right )^{m}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}d x\]
\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \cos \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\left (d \cos {\left (e + f x \right )}\right )^{m}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]
\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \cos \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \cos \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (d\,\cos \left (e+f\,x\right )\right )}^m}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]