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

Optimal. Leaf size=187 \[ \frac{\cos ^2(e+f x) (d \cos (e+f x))^m \left (1-\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right )^{\frac{m+2}{2}} \left (1-\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )^{\frac{m+2}{2}} (a+b \tan (e+f x))^{n+1} F_1\left (n+1;\frac{m+2}{2},\frac{m+2}{2};n+2;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}},\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )}{b f (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.202884, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3515, 3512, 760, 133} \[ \frac{\cos ^2(e+f x) (d \cos (e+f x))^m \left (1-\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}}\right )^{\frac{m+2}{2}} \left (1-\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )^{\frac{m+2}{2}} (a+b \tan (e+f x))^{n+1} F_1\left (n+1;\frac{m+2}{2},\frac{m+2}{2};n+2;\frac{a+b \tan (e+f x)}{a-\sqrt{-b^2}},\frac{a+b \tan (e+f x)}{a+\sqrt{-b^2}}\right )}{b f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rule 760

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[(a + c*x^
2)^p/(e*(1 - (d + e*x)/(d + (e*q)/c))^p*(1 - (d + e*x)/(d - (e*q)/c))^p), Subst[Int[x^m*Simp[1 - x/(d + (e*q)/
c), x]^p*Simp[1 - x/(d - (e*q)/c), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a
*e^2, 0] &&  !IntegerQ[p]

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

Mathematica [C]  time = 22.4844, size = 698, normalized size = 3.73 \[ \frac{2 (d \cos (e+f x))^m (a+b \tan (e+f x))^{n+1} F_1\left (n+1;\frac{m}{2}+1,\frac{m}{2}+1;n+2;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )}{f \left (2 n (b-a \tan (e+f x)) F_1\left (n+1;\frac{m}{2}+1,\frac{m}{2}+1;n+2;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )+2 (n-m) \tan (e+f x) (a+b \tan (e+f x)) F_1\left (n+1;\frac{m}{2}+1,\frac{m}{2}+1;n+2;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )+2 b \sec ^2(e+f x) F_1\left (n+1;\frac{m}{2}+1,\frac{m}{2}+1;n+2;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )+\frac{b (m+2) (n+1) \sec ^2(e+f x) (a+b \tan (e+f x)) \left ((a-i b) F_1\left (n+2;\frac{m}{2}+1,\frac{m}{2}+2;n+3;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )+(a+i b) F_1\left (n+2;\frac{m}{2}+2,\frac{m}{2}+1;n+3;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )\right )}{(n+2) (a-i b) (a+i b)}+\frac{m \sec ^2(e+f x) (a+b \tan (e+f x)) F_1\left (n+1;\frac{m}{2}+1,\frac{m}{2}+1;n+2;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )}{\tan (e+f x)-i}+\frac{m \sec ^2(e+f x) (a+b \tan (e+f x)) F_1\left (n+1;\frac{m}{2}+1,\frac{m}{2}+1;n+2;\frac{a+b \tan (e+f x)}{a-i b},\frac{a+b \tan (e+f x)}{a+i b}\right )}{\tan (e+f x)+i}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*AppellF1[1 + n, 1 + m/2, 1 + m/2, 2 + n, (a + b*Tan[e + f*x])/(a - I*b), (a + b*Tan[e + f*x])/(a + I*b)]*(d
*Cos[e + f*x])^m*(a + b*Tan[e + f*x])^(1 + n))/(f*(2*b*AppellF1[1 + n, 1 + m/2, 1 + m/2, 2 + n, (a + b*Tan[e +
f*x])/(a - I*b), (a + b*Tan[e + f*x])/(a + I*b)]*Sec[e + f*x]^2 + 2*n*AppellF1[1 + n, 1 + m/2, 1 + m/2, 2 + n
, (a + b*Tan[e + f*x])/(a - I*b), (a + b*Tan[e + f*x])/(a + I*b)]*(b - a*Tan[e + f*x]) + (b*(2 + m)*(1 + n)*((
a - I*b)*AppellF1[2 + n, 1 + m/2, 2 + m/2, 3 + n, (a + b*Tan[e + f*x])/(a - I*b), (a + b*Tan[e + f*x])/(a + I*
b)] + (a + I*b)*AppellF1[2 + n, 2 + m/2, 1 + m/2, 3 + n, (a + b*Tan[e + f*x])/(a - I*b), (a + b*Tan[e + f*x])/
(a + I*b)])*Sec[e + f*x]^2*(a + b*Tan[e + f*x]))/((a - I*b)*(a + I*b)*(2 + n)) + 2*(-m + n)*AppellF1[1 + n, 1
+ m/2, 1 + m/2, 2 + n, (a + b*Tan[e + f*x])/(a - I*b), (a + b*Tan[e + f*x])/(a + I*b)]*Tan[e + f*x]*(a + b*Tan
[e + f*x]) + (m*AppellF1[1 + n, 1 + m/2, 1 + m/2, 2 + n, (a + b*Tan[e + f*x])/(a - I*b), (a + b*Tan[e + f*x])/
(a + I*b)]*Sec[e + f*x]^2*(a + b*Tan[e + f*x]))/(-I + Tan[e + f*x]) + (m*AppellF1[1 + n, 1 + m/2, 1 + m/2, 2 +
n, (a + b*Tan[e + f*x])/(a - I*b), (a + b*Tan[e + f*x])/(a + I*b)]*Sec[e + f*x]^2*(a + b*Tan[e + f*x]))/(I +
Tan[e + f*x])))

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Maple [F]  time = 0.239, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \left ( fx+e \right ) \right ) ^{m} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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