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

Optimal. Leaf size=140 \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m F_1\left (\frac{1}{2};1,\frac{m+2}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a f}+\frac{b (d \cos (e+f x))^m \, _2F_1\left (1,-\frac{m}{2};1-\frac{m}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )} \]

[Out]

(b*(d*Cos[e + f*x])^m*Hypergeometric2F1[1, -m/2, 1 - m/2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)])/((a^2 + b^2)*f*m)
+ (AppellF1[1/2, 1, (2 + m)/2, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*(d*Cos[e + f*x])^m*(Sec[e + f*
x]^2)^(m/2)*Tan[e + f*x])/(a*f)

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Rubi [A]  time = 0.210547, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3515, 3512, 757, 429, 444, 68} \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m F_1\left (\frac{1}{2};1,\frac{m+2}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a f}+\frac{b (d \cos (e+f x))^m \, _2F_1\left (1,-\frac{m}{2};1-\frac{m}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(d*Cos[e + f*x])^m*Hypergeometric2F1[1, -m/2, 1 - m/2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)])/((a^2 + b^2)*f*m)
+ (AppellF1[1/2, 1, (2 + m)/2, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*(d*Cos[e + f*x])^m*(Sec[e + f*
x]^2)^(m/2)*Tan[e + f*x])/(a*f)

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 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - (e*x)/(d^2 - e^2*x^2))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 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]

Rubi steps

\begin{align*} \int \frac{(d \cos (e+f x))^m}{a+b \tan (e+f x)} \, dx &=\left ((d \cos (e+f x))^m (d \sec (e+f x))^m\right ) \int \frac{(d \sec (e+f x))^{-m}}{a+b \tan (e+f x)} \, dx\\ &=\frac{\left ((d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}}}{a+x} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{\left ((d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int \left (\frac{a \left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}}}{a^2-x^2}+\frac{x \left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}}}{-a^2+x^2}\right ) \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{\left ((d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int \frac{x \left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}}}{-a^2+x^2} \, dx,x,b \tan (e+f x)\right )}{b f}+\frac{\left (a (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^{-1-\frac{m}{2}}}{a^2-x^2} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac{F_1\left (\frac{1}{2};1,\frac{2+m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan (e+f x)}{a f}+\frac{\left ((d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{b^2}\right )^{-1-\frac{m}{2}}}{-a^2+x} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b f}\\ &=\frac{b (d \cos (e+f x))^m \, _2F_1\left (1,-\frac{m}{2};1-\frac{m}{2};\frac{b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{\left (a^2+b^2\right ) f m}+\frac{F_1\left (\frac{1}{2};1,\frac{2+m}{2};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan (e+f x)}{a f}\\ \end{align*}

Mathematica [C]  time = 14.1019, size = 1132, normalized size = 8.09 \[ \frac{(d \cos (e+f x))^m \left (-b F_1\left (m;\frac{m}{2},\frac{m}{2};m+1;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{m/2} \sec ^2(e+f x)^{-m/2}+b \left (\sec ^2(e+f x)^{-m/2}-1\right )+a m \, _2F_1\left (\frac{1}{2},\frac{m}{2}+1;\frac{3}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)\right )}{f (a+b \tan (e+f x)) \left (-\frac{1}{2} b m F_1\left (m;\frac{m}{2},\frac{m}{2};m+1;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)^{-m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{m/2} \left (\frac{b \sec ^2(e+f x)}{a+b \tan (e+f x)}-\frac{b^2 \sec ^2(e+f x) (\tan (e+f x)-i)}{(a+b \tan (e+f x))^2}\right ) \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{\frac{m}{2}-1}+b m F_1\left (m;\frac{m}{2},\frac{m}{2};m+1;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)^{-m/2} \tan (e+f x) \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{m/2} \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{m/2}-b \sec ^2(e+f x)^{-m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{m/2} \left (-\frac{(a-i b) b m^2 F_1\left (m+1;\frac{m}{2}+1,\frac{m}{2};m+2;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)}{2 (m+1) (a+b \tan (e+f x))^2}-\frac{(a+i b) b m^2 F_1\left (m+1;\frac{m}{2},\frac{m}{2}+1;m+2;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)}{2 (m+1) (a+b \tan (e+f x))^2}\right ) \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{m/2}-\frac{1}{2} b m F_1\left (m;\frac{m}{2},\frac{m}{2};m+1;\frac{a-i b}{a+b \tan (e+f x)},\frac{a+i b}{a+b \tan (e+f x)}\right ) \sec ^2(e+f x)^{-m/2} \left (\frac{b (\tan (e+f x)+i)}{a+b \tan (e+f x)}\right )^{\frac{m}{2}-1} \left (\frac{b \sec ^2(e+f x)}{a+b \tan (e+f x)}-\frac{b^2 \sec ^2(e+f x) (\tan (e+f x)+i)}{(a+b \tan (e+f x))^2}\right ) \left (\frac{b (\tan (e+f x)-i)}{a+b \tan (e+f x)}\right )^{m/2}+a m \, _2F_1\left (\frac{1}{2},\frac{m}{2}+1;\frac{3}{2};-\tan ^2(e+f x)\right ) \sec ^2(e+f x)-b m \sec ^2(e+f x)^{-m/2} \tan (e+f x)+a m \sec ^2(e+f x) \left (\left (\tan ^2(e+f x)+1\right )^{-\frac{m}{2}-1}-\, _2F_1\left (\frac{1}{2},\frac{m}{2}+1;\frac{3}{2};-\tan ^2(e+f x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((d*Cos[e + f*x])^m*(b*(-1 + (Sec[e + f*x]^2)^(-m/2)) + a*m*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]
^2]*Tan[e + f*x] - (b*AppellF1[m, m/2, m/2, 1 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*
x])]*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))
/(Sec[e + f*x]^2)^(m/2)))/(f*(a + b*Tan[e + f*x])*(a*m*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*S
ec[e + f*x]^2 - (b*m*Tan[e + f*x])/(Sec[e + f*x]^2)^(m/2) + (b*m*AppellF1[m, m/2, m/2, 1 + m, (a - I*b)/(a + b
*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*Tan[e + f*x]*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m
/2)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))/(Sec[e + f*x]^2)^(m/2) - (b*((b*(-I + Tan[e + f*x]))/
(a + b*Tan[e + f*x]))^(m/2)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*(-((a - I*b)*b*m^2*AppellF1[1
+ m, 1 + m/2, m/2, 2 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*Sec[e + f*x]^2)/(2*(
1 + m)*(a + b*Tan[e + f*x])^2) - ((a + I*b)*b*m^2*AppellF1[1 + m, m/2, 1 + m/2, 2 + m, (a - I*b)/(a + b*Tan[e
+ f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*Sec[e + f*x]^2)/(2*(1 + m)*(a + b*Tan[e + f*x])^2)))/(Sec[e + f*x]^2)
^(m/2) - (b*m*AppellF1[m, m/2, m/2, 1 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*((b
*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(-1 + m/2)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*(-(
(b^2*Sec[e + f*x]^2*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x])^2) + (b*Sec[e + f*x]^2)/(a + b*Tan[e + f*x])))/(
2*(Sec[e + f*x]^2)^(m/2)) - (b*m*AppellF1[m, m/2, m/2, 1 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b
*Tan[e + f*x])]*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*
x]))^(-1 + m/2)*(-((b^2*Sec[e + f*x]^2*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x])^2) + (b*Sec[e + f*x]^2)/(a + b
*Tan[e + f*x])))/(2*(Sec[e + f*x]^2)^(m/2)) + a*m*Sec[e + f*x]^2*(-Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e
+ f*x]^2] + (1 + Tan[e + f*x]^2)^(-1 - m/2))))

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Maple [F]  time = 0.216, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\cos \left ( fx+e \right ) \right ) ^{m}}{a+b\tan \left ( fx+e \right ) }}\, 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)),x)

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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