3.7.99 \(\int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx\) [699]
Optimal. Leaf size=227 \[ \frac {2 a b (d \cos (e+f x))^m \, _2F_1\left (2,-\frac {m}{2};1-\frac {m}{2};\frac {b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{\left (a^2+b^2\right )^2 f m}+\frac {F_1\left (\frac {1}{2};2,\frac {2+m}{2};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan (e+f x)}{a^2 f}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {2+m}{2};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan ^3(e+f x)}{3 a^4 f} \]
[Out]
2*a*b*(d*cos(f*x+e))^m*hypergeom([2, -1/2*m],[1-1/2*m],b^2*sec(f*x+e)^2/(a^2+b^2))/(a^2+b^2)^2/f/m+AppellF1(1/
2,2,1+1/2*m,3/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*cos(f*x+e))^m*(sec(f*x+e)^2)^(1/2*m)*tan(f*x+e)/a^2/f+1
/3*b^2*AppellF1(3/2,2,1+1/2*m,5/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*cos(f*x+e))^m*(sec(f*x+e)^2)^(1/2*m)*
tan(f*x+e)^3/a^4/f
________________________________________________________________________________________
Rubi [A]
time = 0.20, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3596, 3593,
771, 440, 455, 70, 524} \begin {gather*} \frac {\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m F_1\left (\frac {1}{2};2,\frac {m+2}{2};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f}+\frac {2 a b (d \cos (e+f x))^m \, _2F_1\left (2,-\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 )^2}+\frac {b^2 \tan ^3(e+f x) \sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m F_1\left (\frac {3}{2};2,\frac {m+2}{2};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(d*Cos[e + f*x])^m/(a + b*Tan[e + f*x])^2,x]
[Out]
(2*a*b*(d*Cos[e + f*x])^m*Hypergeometric2F1[2, -1/2*m, 1 - m/2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)])/((a^2 + b^2
)^2*f*m) + (AppellF1[1/2, 2, (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*(Se
c[e + f*x]^2)^(m/2)*Tan[e + f*x])/(a^2*f) + (b^2*AppellF1[3/2, 2, (2 + m)/2, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -T
an[e + f*x]^2]*(d*Cos[e + f*x])^m*(Sec[e + f*x]^2)^(m/2)*Tan[e + f*x]^3)/(3*a^4*f)
Rule 70
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && IntegerQ[n]
Rule 440
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 455
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 524
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rule 771
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 3593
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 3596
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]
Rubi steps
\begin {align*} \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, 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))^2} \, dx\\ &=\frac {\left ((d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}}}{(a+x)^2} \, 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 ) \text {Subst}\left (\int \left (\frac {a^2 \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}}}{\left (a^2-x^2\right )^2}-\frac {2 a x \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}}}{\left (a^2-x^2\right )^2}+\frac {x^2 \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}}}{\left (-a^2+x^2\right )^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 ) \text {Subst}\left (\int \frac {x^2 \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}}}{\left (-a^2+x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}-\frac {\left (2 a (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \text {Subst}\left (\int \frac {x \left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}}}{\left (a^2-x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}+\frac {\left (a^2 (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {x^2}{b^2}\right )^{-1-\frac {m}{2}}}{\left (a^2-x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac {F_1\left (\frac {1}{2};2,\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^2 f}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {2+m}{2};\frac {5}{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 ^3(e+f x)}{3 a^4 f}-\frac {\left (a (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {x}{b^2}\right )^{-1-\frac {m}{2}}}{\left (a^2-x\right )^2} \, dx,x,b^2 \tan ^2(e+f x)\right )}{b f}\\ &=\frac {2 a b (d \cos (e+f x))^m \, _2F_1\left (2,-\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 )^2 f m}+\frac {F_1\left (\frac {1}{2};2,\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^2 f}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {2+m}{2};\frac {5}{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 ^3(e+f x)}{3 a^4 f}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 17.26, size = 2502, normalized size = 11.02 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*Cos[e + f*x])^m/(a + b*Tan[e + f*x])^2,x]
[Out]
((d*Cos[e + f*x])^m*((2*a*b*(-1 + (Sec[e + f*x]^2)^(-1/2*m)))/m + a^2*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Ta
n[e + f*x]^2]*Tan[e + f*x] - b^2*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x] - (2*a*b*A
ppellF1[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))/(m*(Sec[e + f*x]^2)^(m
/2)) - (b*(a^2 + b^2)*AppellF1[1 + m, m/2, m/2, 2 + 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))/((1 + m)*(Sec[e + f*x]^2)^(m/2)*(a + b*Tan[e + f*x]))))/(f*(a + b*Tan[e + f*x])^2*(a^2*Hypergeometric2F1[
1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Sec[e + f*x]^2 - b^2*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]
*Sec[e + f*x]^2 - (2*a*b*Tan[e + f*x])/(Sec[e + f*x]^2)^(m/2) + (2*a*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])]*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^2*(a^2 + b^2)*Appel
lF1[1 + m, 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)^(
1 - m/2)*((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))/((1 + m)*(a + b*Tan[e + f*x])^2) + (b*(a^2 + b^2)*m*AppellF1[1 + m, m/2, m/2, 2 + 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))/((1 + m)*(Sec[e + f*x]^2)^(m/2)*(a + b*Tan[e + f*x])) - (
2*a*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)
*(-1/2*((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)/((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*Ta
n[e + f*x])^2)))/(m*(Sec[e + f*x]^2)^(m/2)) - (b*(a^2 + b^2)*((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)*(-1/2*((a - I*b)*b*m*(1 + m)*AppellF1[2 + m, 1 + m/2,
m/2, 3 + 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 + m)*(a + b*Ta
n[e + f*x])^2) - ((a + I*b)*b*m*(1 + m)*AppellF1[2 + m, m/2, 1 + m/2, 3 + 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*(2 + m)*(a + b*Tan[e + f*x])^2)))/((1 + m)*(Sec[e + f*x]^2)^
(m/2)*(a + b*Tan[e + f*x])) - (a*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]))^(-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])))/(Sec[e + f*x]^2)^(m/2) - (b*(a^2 + b^2)*m*AppellF1[1 + m, m/2, m/2, 2 + 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*(1 + m)*(Sec[e + f*x]^2)^(m/2)*(a + b*Tan[e + f*x])) - (
a*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]))^(-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])))/(Sec[e + f*x
]^2)^(m/2) - (b*(a^2 + b^2)*m*AppellF1[1 + m, m/2, m/2, 2 + 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*(1 + m)*(Sec[e + f*x]^2)^(m/2)*(a + b*Tan[e + f*x])) + a^2*Sec[e + f*x]^2*(-Hypergeometri
c2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2] + (1 + Tan[e + f*x]^2)^(-1 - m/2)) - b^2*Sec[e + f*x]^2*(-Hypergeomet
ric2F1[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 = 1.29, size = 0, normalized size = 0.00 \[\int \frac {\left (d \cos \left (f x +e \right )\right )^{m}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x)
[Out]
int((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x)
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
[Out]
integrate((d*cos(f*x + e))^m/(b*tan(f*x + e) + a)^2, x)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
[Out]
integral((d*cos(f*x + e))^m/(b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2), x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d \cos {\left (e + f x \right )}\right )^{m}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*cos(f*x+e))**m/(a+b*tan(f*x+e))**2,x)
[Out]
Integral((d*cos(e + f*x))**m/(a + b*tan(e + f*x))**2, x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="giac")
[Out]
integrate((d*cos(f*x + e))^m/(b*tan(f*x + e) + a)^2, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d\,\cos \left (e+f\,x\right )\right )}^m}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*cos(e + f*x))^m/(a + b*tan(e + f*x))^2,x)
[Out]
int((d*cos(e + f*x))^m/(a + b*tan(e + f*x))^2, x)
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