3.1309 \(\int \frac{(a+b \tan (e+f x))^m}{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=223 \[ \frac{d^2 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right ) (b c-a d)}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)}-\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (-d+i c)} \]

[Out]

(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)) - (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)*(I*c - d)*f*(1 + m)) + (d^2*Hypergeometric2F1[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))

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Rubi [A]  time = 0.374174, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3574, 3539, 3537, 68, 3634} \[ \frac{d^2 (a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b \tan (e+f x))}{b c-a d}\right )}{f (m+1) \left (c^2+d^2\right ) (b c-a d)}+\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{2 f (m+1) (b+i a) (c-i d)}-\frac{(a+b \tan (e+f x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{2 f (m+1) (a+i b) (-d+i c)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x]),x]

[Out]

(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)) - (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)*(I*c - d)*f*(1 + m)) + (d^2*Hypergeometric2F1[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 3574

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/
(c^2 + d^2), Int[(a + b*Tan[e + f*x])^m*(c - d*Tan[e + f*x]), x], x] + Dist[d^2/(c^2 + d^2), Int[((a + b*Tan[e
+ f*x])^m*(1 + Tan[e + f*x]^2))/(c + d*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 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}{c+d \tan (e+f x)} \, dx &=\frac{\int (a+b \tan (e+f x))^m (c-d \tan (e+f x)) \, dx}{c^2+d^2}+\frac{d^2 \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{\int (1+i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c-i d)}+\frac{\int (1-i \tan (e+f x)) (a+b \tan (e+f x))^m \, dx}{2 (c+i d)}+\frac{d^2 \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{d^2 \, _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{\operatorname{Subst}\left (\int \frac{(a+i b x)^m}{-1+x} \, dx,x,-i \tan (e+f x)\right )}{2 (i c-d) f}-\frac{\operatorname{Subst}\left (\int \frac{(a-i b x)^m}{-1+x} \, dx,x,i \tan (e+f x)\right )}{2 (i c+d) f}\\ &=\frac{\, _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{\, _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) (i c-d) f (1+m)}+\frac{d^2 \, _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 = 0.42618, size = 178, normalized size = 0.8 \[ \frac{(a+b \tan (e+f x))^{m+1} \left (-\frac{2 d^2 \, _2F_1\left (1,m+1;m+2;\frac{d (a+b \tan (e+f x))}{a d-b c}\right )}{\left (c^2+d^2\right ) (a d-b c)}+\frac{\, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a-i b}\right )}{(b+i a) (c-i d)}+\frac{i \, _2F_1\left (1,m+1;m+2;\frac{a+b \tan (e+f x)}{a+i b}\right )}{(a+i b) (c+i d)}\right )}{2 f (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^m/(c + d*Tan[e + f*x]),x]

[Out]

((Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a - I*b)]/((I*a + b)*(c - I*d)) + (I*Hypergeometric
2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)])/((a + I*b)*(c + I*d)) - (2*d^2*Hypergeometric2F1[1, 1 +
m, 2 + m, (d*(a + b*Tan[e + f*x]))/(-(b*c) + a*d)])/((-(b*c) + a*d)*(c^2 + d^2)))*(a + b*Tan[e + f*x])^(1 + m)
)/(2*f*(1 + m))

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Maple [F]  time = 0.26, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{m}}{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/(c+d*tan(f*x+e)),x)

[Out]

int((a+b*tan(f*x+e))^m/(c+d*tan(f*x+e)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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/(c+d*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\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/(c+d*tan(f*x+e)),x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{m}}{c + d \tan{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**m/(c+d*tan(f*x+e)),x)

[Out]

Integral((a + b*tan(e + f*x))**m/(c + d*tan(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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/(c+d*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e) + a)^m/(d*tan(f*x + e) + c), x)