3.169 \(\int \frac{(a+b \tan (e+f x))^m (A+B \tan (e+f x)+C \tan ^2(e+f x))}{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=258 \[ \frac{\left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right ) (b c-a d)}-\frac{(i A+B-i C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b) (c-i d)}-\frac{(A+i B-C) (a+b \tan (e+f x))^{m+1} \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (c+i d)} \]

[Out]

-((I*A + B - I*C)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 +
m))/(2*(a - I*b)*(c - I*d)*f*(1 + m)) - ((A + I*B - C)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x]
)/(a + I*b)]*(a + b*Tan[e + f*x])^(1 + m))/(2*(I*a - b)*(c + I*d)*f*(1 + m)) + ((c^2*C - B*c*d + A*d^2)*Hyperg
eometric2F1[1, 1 + m, 2 + m, -((d*(a + b*Tan[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))

________________________________________________________________________________________

Rubi [A]  time = 0.482221, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3653, 3539, 3537, 68, 3634} \[ \frac{\left (A d^2-B c d+c^2 C\right ) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right ) (b c-a d)}-\frac{(i A+B-i C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (a-i b) (c-i d)}-\frac{(A+i B-C) (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (-b+i a) (c+i d)} \]

Antiderivative was successfully verified.

[In]

Int[((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]

[Out]

-((I*A + B - I*C)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 +
m))/(2*(a - I*b)*(c - I*d)*f*(1 + m)) - ((A + I*B - C)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x]
)/(a + I*b)]*(a + b*Tan[e + f*x])^(1 + m))/(2*(I*a - b)*(c + I*d)*f*(1 + m)) + ((c^2*C - B*c*d + A*d^2)*Hyperg
eometric2F1[1, 1 + m, 2 + m, -((d*(a + b*Tan[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))

Rule 3653

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] :> Dist[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] + Dist[(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]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(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]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^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]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3634

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[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]

Rubi steps

\begin{align*} \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{\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 (c^2 C-B c d+A d^2\right ) \int \frac{(a+b \tan (e+f x))^m \left (1+\tan ^2(e+f x)\right )}{c+d \tan (e+f x)} \, dx}{c^2+d^2}\\ &=\frac{(A-i B-C) \int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c-i d)}+\frac{(A+i B-C) \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c+i d)}+\frac{\left (c^2 C-B c d+A d^2\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{c+d x} \, dx,x,\tan (e+f x)\right )}{\left (c^2+d^2\right ) f}\\ &=\frac{\left (c^2 C-B c d+A d^2\right ) \, _2F_1\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)}+\frac{(i A+B-i C) \operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 (c-i d) f}-\frac{(i (A+i B-C)) \operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 (c+i d) f}\\ &=-\frac{(i A+B-i C) \, _2F_1\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) \, _2F_1\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 ) \, _2F_1\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)}\\ \end{align*}

Mathematica [A]  time = 0.998107, size = 204, normalized size = 0.79 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (\frac{2 \left (A d^2-B c d+c^2 C\right ) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{d (a+b \tan (e+f x))}{a d-b c}\right )}{b c-a d}+\frac{(d-i c) (A-i B-C) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a-i b}\right )}{a-i b}+\frac{(d+i c) (A+i B-C) \text{Hypergeometric2F1}\left (1,m+1,m+2,\frac{a+b \tan (e+f x)}{a+i b}\right )}{a+i b}\right )}{2 f (m+1) \left (c^2+d^2\right )} \]

Antiderivative was successfully verified.

[In]

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]

[Out]

((((A - I*B - C)*((-I)*c + d)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)])/(a - I*b) +
((A + I*B - C)*(I*c + d)*Hypergeometric2F1[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))

________________________________________________________________________________________

Maple [F]  time = 0.543, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m} \left ( A+B\tan \left ( fx+e \right ) +C \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{c+d\tan \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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)

[Out]

int((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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\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}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)