3.411 \(\int (d \tan (e+f x))^n (a+a \tan (e+f x))^m \, dx\)

Optimal. Leaf size=161 \[ \frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),-i \tan (e+f x))}{2 d f (n+1)}+\frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),i \tan (e+f x))}{2 d f (n+1)} \]

[Out]

(AppellF1[1 + n, -m, 1, 2 + n, -Tan[e + f*x], (-I)*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n)*(a + a*Tan[e + f*x])
^m)/(2*d*f*(1 + n)*(1 + Tan[e + f*x])^m) + (AppellF1[1 + n, -m, 1, 2 + n, -Tan[e + f*x], I*Tan[e + f*x]]*(d*Ta
n[e + f*x])^(1 + n)*(a + a*Tan[e + f*x])^m)/(2*d*f*(1 + n)*(1 + Tan[e + f*x])^m)

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Rubi [A]  time = 0.14935, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3575, 912, 135, 133} \[ \frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),-i \tan (e+f x))}{2 d f (n+1)}+\frac{(\tan (e+f x)+1)^{-m} (a \tan (e+f x)+a)^m (d \tan (e+f x))^{n+1} F_1(n+1;-m,1;n+2;-\tan (e+f x),i \tan (e+f x))}{2 d f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[1 + n, -m, 1, 2 + n, -Tan[e + f*x], (-I)*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n)*(a + a*Tan[e + f*x])
^m)/(2*d*f*(1 + n)*(1 + Tan[e + f*x])^m) + (AppellF1[1 + n, -m, 1, 2 + n, -Tan[e + f*x], I*Tan[e + f*x]]*(d*Ta
n[e + f*x])^(1 + n)*(a + a*Tan[e + f*x])^m)/(2*d*f*(1 + n)*(1 + Tan[e + f*x])^m)

Rule 3575

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n)/(1 + ff^2*x^2), x]
, x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 912

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int (d \tan (e+f x))^n (a+a \tan (e+f x))^m \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(d x)^n (a+a x)^m}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i (d x)^n (a+a x)^m}{2 (i-x)}+\frac{i (d x)^n (a+a x)^m}{2 (i+x)}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{(d x)^n (a+a x)^m}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{i \operatorname{Subst}\left (\int \frac{(d x)^n (a+a x)^m}{i+x} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{\left (i (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{(d x)^n (1+x)^m}{i-x} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left (i (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m\right ) \operatorname{Subst}\left (\int \frac{(d x)^n (1+x)^m}{i+x} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac{F_1(1+n;-m,1;2+n;-\tan (e+f x),-i \tan (e+f x)) (d \tan (e+f x))^{1+n} (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m}{2 d f (1+n)}+\frac{F_1(1+n;-m,1;2+n;-\tan (e+f x),i \tan (e+f x)) (d \tan (e+f x))^{1+n} (1+\tan (e+f x))^{-m} (a+a \tan (e+f x))^m}{2 d f (1+n)}\\ \end{align*}

Mathematica [F]  time = 0.677661, size = 0, normalized size = 0. \[ \int (d \tan (e+f x))^n (a+a \tan (e+f x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.615, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+a\tan \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^n*(a+a*tan(f*x+e))^m,x)

[Out]

int((d*tan(f*x+e))^n*(a+a*tan(f*x+e))^m,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n*(a+a*tan(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((a*tan(f*x + e) + a)^m*(d*tan(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n*(a+a*tan(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((a*tan(f*x + e) + a)^m*(d*tan(f*x + e))^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\tan{\left (e + f x \right )} + 1\right )\right )^{m} \left (d \tan{\left (e + f x \right )}\right )^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**n*(a+a*tan(f*x+e))**m,x)

[Out]

Integral((a*(tan(e + f*x) + 1))**m*(d*tan(e + f*x))**n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n*(a+a*tan(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((a*tan(f*x + e) + a)^m*(d*tan(f*x + e))^n, x)