3.644 \(\int \frac {(d \sec (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx\)
Optimal. Leaf size=227 \[ \frac {\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac {1}{2};2,1-\frac {m}{2};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f}-\frac {2 a b (d \sec (e+f x))^m \, _2F_1\left (2,\frac {m}{2};\frac {m+2}{2};\frac {b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )^2}+\frac {b^2 \tan ^3(e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac {3}{2};2,1-\frac {m}{2};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f} \]
[Out]
-2*a*b*hypergeom([2, 1/2*m],[1+1/2*m],b^2*sec(f*x+e)^2/(a^2+b^2))*(d*sec(f*x+e))^m/(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*sec(f*x+e))^m*tan(f*x+e)/a^2/f/((sec(f*x+e)^2)^(1/2*m))
+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*sec(f*x+e))^m*tan(f*x+e)^3/a^4/f/((
sec(f*x+e)^2)^(1/2*m))
________________________________________________________________________________________
Rubi [A] time = 0.20, antiderivative size = 227, normalized size of antiderivative = 1.00,
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 =
{3512, 757, 429, 444, 68, 510} \[ \frac {b^2 \tan ^3(e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac {3}{2};2,1-\frac {m}{2};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f}+\frac {\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \sec (e+f x))^m F_1\left (\frac {1}{2};2,1-\frac {m}{2};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f}-\frac {2 a b (d \sec (e+f x))^m \, _2F_1\left (2,\frac {m}{2};\frac {m+2}{2};\frac {b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{f m \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^m/(a + b*Tan[e + f*x])^2,x]
[Out]
(-2*a*b*Hypergeometric2F1[2, m/2, (2 + m)/2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*(d*Sec[e + f*x])^m)/((a^2 + b^2
)^2*f*m) + (AppellF1[1/2, 2, 1 - m/2, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*(d*Sec[e + f*x])^m*Tan[e
+ f*x])/(a^2*f*(Sec[e + f*x]^2)^(m/2)) + (b^2*AppellF1[3/2, 2, 1 - m/2, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e
+ f*x]^2]*(d*Sec[e + f*x])^m*Tan[e + f*x]^3)/(3*a^4*f*(Sec[e + f*x]^2)^(m/2))
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]
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 510
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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 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 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]
Rubi steps
\begin {align*} \int \frac {(d \sec (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx &=\frac {\left ((d \sec (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)^2} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac {\left ((d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname {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 \sec (e+f x))^m \sec ^2(e+f x)^{-m/2}\right ) \operatorname {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 \sec (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}}}{\left (a^2-x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}+\frac {\left (a^2 (d \sec (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}}}{\left (a^2-x^2\right )^2} \, dx,x,b \tan (e+f x)\right )}{b f}\\ &=\frac {F_1\left (\frac {1}{2};2,1-\frac {m}{2};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (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};2,1-\frac {m}{2};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan ^3(e+f x)}{3 a^4 f}-\frac {\left (a (d \sec (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}}}{\left (a^2-x\right )^2} \, dx,x,b^2 \tan ^2(e+f x)\right )}{b f}\\ &=-\frac {2 a b \, _2F_1\left (2,\frac {m}{2};\frac {2+m}{2};\frac {b^2 \sec ^2(e+f x)}{a^2+b^2}\right ) (d \sec (e+f x))^m}{\left (a^2+b^2\right )^2 f m}+\frac {F_1\left (\frac {1}{2};2,1-\frac {m}{2};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (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};2,1-\frac {m}{2};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (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] time = 18.52, size = 2453, normalized size = 10.81 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*Sec[e + f*x])^m/(a + b*Tan[e + f*x])^2,x]
[Out]
((d*Sec[e + f*x])^m*((-2*a*b*(-1 + (Sec[e + f*x]^2)^(m/2)))/m + (a^2 - b^2)*Hypergeometric2F1[1/2, 1 - m/2, 3/
2, -Tan[e + f*x]^2]*Tan[e + f*x] + (2*a*b*AppellF1[-m, -1/2*m, -1/2*m, 1 - m, (a - I*b)/(a + b*Tan[e + f*x]),
(a + I*b)/(a + b*Tan[e + f*x])]*(Sec[e + f*x]^2)^(m/2))/(m*((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)) + (b*(a^2 + b^2)*AppellF1[1 - m, -1/2*m, -1/2*m, 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)^(m/2))/((-1 + m)*((b*(-I + T
an[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 + b*Tan[e + f
*x]))))/(f*(a + b*Tan[e + f*x])^2*((a^2 - b^2)*Hypergeometric2F1[1/2, 1 - m/2, 3/2, -Tan[e + f*x]^2]*Sec[e + f
*x]^2 - 2*a*b*(Sec[e + f*x]^2)^(m/2)*Tan[e + f*x] + (2*a*b*AppellF1[-m, -1/2*m, -1/2*m, 1 - m, (a - I*b)/(a +
b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*(Sec[e + f*x]^2)^(m/2)*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)) - (b^2*(a^2 + b^2)*AppellF1
[1 - m, -1/2*m, -1/2*m, 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))/((-1 + m)*((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 + b*Tan[e + f*x])^2) + (b*(a^2 + b^2)*m*AppellF1[1 - m, -1/2*m, -1/2*m, 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)^(m/2)*Tan[e + f*x])/((-1 + m)*((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 + b*Tan[e +
f*x])) + (2*a*b*(Sec[e + f*x]^2)^(m/2)*(-1/2*((a - I*b)*b*m^2*AppellF1[1 - m, 1 - m/2, -1/2*m, 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, -1/2*m, 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)))/(m*((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)) + (b*(a^2 + b^2)*(Sec[e + f*x]^2)^(m/2)*(((a - I*
b)*b*(1 - m)*m*AppellF1[2 - m, 1 - m/2, -1/2*m, 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) + ((a + I*b)*b*(1 - m)*m*AppellF1[2 - m, -1/2*m, 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)*((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 + b*Tan[e + f*x])) - (a*b*AppellF1[-m, -1/2*m, -1/2*m, 1 - m, (a - I*b)/(a + b*T
an[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*(Sec[e + f*x]^2)^(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])))/((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2) - (b*(a^2 + b^2)*m*AppellF1[1 - m, -1/2
*m, -1/2*m, 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)^(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)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x
]))^(m/2)*(a + b*Tan[e + f*x])) - (a*b*AppellF1[-m, -1/2*m, -1/2*m, 1 - m, (a - I*b)/(a + b*Tan[e + f*x]), (a
+ I*b)/(a + b*Tan[e + f*x])]*(Sec[e + f*x]^2)^(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])))/
((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2) - (b*(a^2 + b^2)*m*AppellF1[1 - m, -1/2*m, -1/2*m, 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)^(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)*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*(a + b*Ta
n[e + f*x])) + (a^2 - b^2)*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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fricas [F] time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d \sec \left (f x + e\right )\right )^{m}}{b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
[Out]
integral((d*sec(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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^m/(b*tan(f*x + e) + a)^2, x)
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maple [F] time = 1.34, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x +e \right )\right )^{m}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^m/(a+b*tan(f*x+e))^2,x)
[Out]
int((d*sec(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 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
[Out]
integrate((d*sec(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 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]
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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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{m}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))**m/(a+b*tan(f*x+e))**2,x)
[Out]
Integral((d*sec(e + f*x))**m/(a + b*tan(e + f*x))**2, x)
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