Integrand size = 45, antiderivative size = 258 \[ \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx=-\frac {(i A+B-i C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a-i b) (c-i d) f (1+m)}-\frac {(A+i B-C) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (i a-b) (c+i d) f (1+m)}+\frac {\left (c^2 C-B c d+A d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d (a+b \tan (e+f x))}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{(b c-a d) \left (c^2+d^2\right ) f (1+m)} \]
-1/2*(I*A+B-I*C)*hypergeom([1, 1+m],[2+m],(a+b*tan(f*x+e))/(a-I*b))*(a+b*t an(f*x+e))^(1+m)/(a-I*b)/(c-I*d)/f/(1+m)-1/2*(A+I*B-C)*hypergeom([1, 1+m], [2+m],(a+b*tan(f*x+e))/(a+I*b))*(a+b*tan(f*x+e))^(1+m)/(I*a-b)/(c+I*d)/f/( 1+m)+(A*d^2-B*c*d+C*c^2)*hypergeom([1, 1+m],[2+m],-d*(a+b*tan(f*x+e))/(-a* d+b*c))*(a+b*tan(f*x+e))^(1+m)/(-a*d+b*c)/(c^2+d^2)/f/(1+m)
Time = 1.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.79 \[ \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx=\frac {\left (\frac {(A-i B-C) (-i c+d) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a-i b}\right )}{a-i b}+\frac {(A+i B-C) (i c+d) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a+i b}\right )}{a+i b}+\frac {2 \left (c^2 C-B c d+A d^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {d (a+b \tan (e+f x))}{-b c+a d}\right )}{b c-a d}\right ) (a+b \tan (e+f x))^{1+m}}{2 \left (c^2+d^2\right ) f (1+m)} \]
Integrate[((a + b*Tan[e + f*x])^m*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2)) /(c + d*Tan[e + f*x]),x]
((((A - I*B - C)*((-I)*c + d)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Ta n[e + f*x])/(a - I*b)])/(a - I*b) + ((A + I*B - C)*(I*c + d)*Hypergeometri c2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)])/(a + I*b) + (2*(c^2 *C - B*c*d + A*d^2)*Hypergeometric2F1[1, 1 + m, 2 + m, (d*(a + b*Tan[e + f *x]))/(-(b*c) + a*d)])/(b*c - a*d))*(a + b*Tan[e + f*x])^(1 + m))/(2*(c^2 + d^2)*f*(1 + m))
Time = 1.11 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4136, 3042, 4022, 3042, 4020, 25, 78, 4117, 78}
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.
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan (e+f x)^2\right )}{c+d \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 4136 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m \left (\tan ^2(e+f x)+1\right )}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {\int (a+b \tan (e+f x))^m (A c-C c+B d+(B c-(A-C) d) \tan (e+f x))dx}{c^2+d^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (a+b \tan (e+f x))^m (A c-C c+B d+(B c-(A-C) d) \tan (e+f x))dx}{c^2+d^2}+\frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m \left (\tan (e+f x)^2+1\right )}{c+d \tan (e+f x)}dx}{c^2+d^2}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m \left (\tan (e+f x)^2+1\right )}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {\frac {1}{2} (c-i d) (A+i B-C) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^mdx+\frac {1}{2} (c+i d) (A-i B-C) \int (i \tan (e+f x)+1) (a+b \tan (e+f x))^mdx}{c^2+d^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m \left (\tan (e+f x)^2+1\right )}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {\frac {1}{2} (c-i d) (A+i B-C) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^mdx+\frac {1}{2} (c+i d) (A-i B-C) \int (i \tan (e+f x)+1) (a+b \tan (e+f x))^mdx}{c^2+d^2}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m \left (\tan (e+f x)^2+1\right )}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {\frac {i (c+i d) (A-i B-C) \int -\frac {(a+b \tan (e+f x))^m}{1-i \tan (e+f x)}d(i \tan (e+f x))}{2 f}-\frac {i (c-i d) (A+i B-C) \int -\frac {(a+b \tan (e+f x))^m}{i \tan (e+f x)+1}d(-i \tan (e+f x))}{2 f}}{c^2+d^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m \left (\tan (e+f x)^2+1\right )}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {\frac {i (c-i d) (A+i B-C) \int \frac {(a+b \tan (e+f x))^m}{i \tan (e+f x)+1}d(-i \tan (e+f x))}{2 f}-\frac {i (c+i d) (A-i B-C) \int \frac {(a+b \tan (e+f x))^m}{1-i \tan (e+f x)}d(i \tan (e+f x))}{2 f}}{c^2+d^2}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m \left (\tan (e+f x)^2+1\right )}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {\frac {i (c-i d) (A+i B-C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}-\frac {i (c+i d) (A-i B-C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}}{c^2+d^2}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) \int \frac {(a+b \tan (e+f x))^m}{c+d \tan (e+f x)}d\tan (e+f x)}{f \left (c^2+d^2\right )}+\frac {\frac {i (c-i d) (A+i B-C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}-\frac {i (c+i d) (A-i B-C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}}{c^2+d^2}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {\left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right ) (b c-a d)}+\frac {\frac {i (c-i d) (A+i B-C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b)}-\frac {i (c+i d) (A-i B-C) (a+b \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b)}}{c^2+d^2}\) |
((c^2*C - B*c*d + A*d^2)*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*(a + b*Ta n[e + f*x]))/(b*c - a*d))]*(a + b*Tan[e + f*x])^(1 + m))/((b*c - a*d)*(c^2 + d^2)*f*(1 + m)) + (((-1/2*I)*(A - I*B - C)*(c + I*d)*Hypergeometric2F1[ 1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 + m))/((a - I*b)*f*(1 + m)) + ((I/2)*(A + I*B - C)*(c - I*d)*Hypergeometric 2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)]*(a + b*Tan[e + f*x])^ (1 + m))/((a + I*b)*f*(1 + m)))/(c^2 + d^2)
3.2.69.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^ n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ (A*b^2 - a*b*B + a^2*C)/(a^2 + b^2) Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & & !GtQ[n, 0] && !LeQ[n, -1]
\[\int \frac {\left (a +b \tan \left (f x +e \right )\right )^{m} \left (A +B \tan \left (f x +e \right )+C \tan \left (f x +e \right )^{2}\right )}{c +d \tan \left (f x +e \right )}d x\]
\[ \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx=\int { \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{d \tan \left (f x + e\right ) + c} \,d x } \]
integrate((a+b*tan(f*x+e))^m*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ e)),x, algorithm="fricas")
integral((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)^m/(d *tan(f*x + e) + c), x)
\[ \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx=\int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{m} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{c + d \tan {\left (e + f x \right )}}\, dx \]
Integral((a + b*tan(e + f*x))**m*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/ (c + d*tan(e + f*x)), x)
\[ \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx=\int { \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{d \tan \left (f x + e\right ) + c} \,d x } \]
integrate((a+b*tan(f*x+e))^m*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ e)),x, algorithm="maxima")
integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)^m/( d*tan(f*x + e) + c), x)
\[ \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx=\int { \frac {{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{m}}{d \tan \left (f x + e\right ) + c} \,d x } \]
integrate((a+b*tan(f*x+e))^m*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(c+d*tan(f*x+ e)),x, algorithm="giac")
integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(b*tan(f*x + e) + a)^m/( d*tan(f*x + e) + c), x)
Timed out. \[ \int \frac {(a+b \tan (e+f x))^m \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,\left (C\,{\mathrm {tan}\left (e+f\,x\right )}^2+B\,\mathrm {tan}\left (e+f\,x\right )+A\right )}{c+d\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]