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

Optimal. Leaf size=179 \[ \frac {(d \tan (e+f x))^{n+1} (a+b \tan (e+f x))^m \left (\frac {b \tan (e+f x)}{a}+1\right )^{-m} F_1\left (n+1;-m,1;n+2;-\frac {b \tan (e+f x)}{a},-i \tan (e+f x)\right )}{2 d f (n+1)}+\frac {(d \tan (e+f x))^{n+1} (a+b \tan (e+f x))^m \left (\frac {b \tan (e+f x)}{a}+1\right )^{-m} F_1\left (n+1;-m,1;n+2;-\frac {b \tan (e+f x)}{a},i \tan (e+f x)\right )}{2 d f (n+1)} \]

[Out]

1/2*AppellF1(1+n,1,-m,2+n,-I*tan(f*x+e),-b*tan(f*x+e)/a)*(d*tan(f*x+e))^(1+n)*(a+b*tan(f*x+e))^m/d/f/(1+n)/((1
+b*tan(f*x+e)/a)^m)+1/2*AppellF1(1+n,1,-m,2+n,I*tan(f*x+e),-b*tan(f*x+e)/a)*(d*tan(f*x+e))^(1+n)*(a+b*tan(f*x+
e))^m/d/f/(1+n)/((1+b*tan(f*x+e)/a)^m)

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Rubi [A]  time = 0.18, antiderivative size = 179, normalized size of antiderivative = 1.00, 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} \[ \frac {(d \tan (e+f x))^{n+1} (a+b \tan (e+f x))^m \left (\frac {b \tan (e+f x)}{a}+1\right )^{-m} F_1\left (n+1;-m,1;n+2;-\frac {b \tan (e+f x)}{a},-i \tan (e+f x)\right )}{2 d f (n+1)}+\frac {(d \tan (e+f x))^{n+1} (a+b \tan (e+f x))^m \left (\frac {b \tan (e+f x)}{a}+1\right )^{-m} F_1\left (n+1;-m,1;n+2;-\frac {b \tan (e+f x)}{a},i \tan (e+f x)\right )}{2 d f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 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]

Rubi steps

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

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Mathematica [F]  time = 1.51, size = 0, normalized size = 0.00 \[ \int (d \tan (e+f x))^n (a+b \tan (e+f x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.87, size = 0, normalized size = 0.00 \[ \int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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