3.7.42 \(\int (d \sec (e+f x))^m (a+b \tan (e+f x)) \, dx\) [642]
Optimal. Leaf size=93 \[ \frac {b (d \sec (e+f x))^m}{f m}-\frac {a d \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{-1+m} \sin (e+f x)}{f (1-m) \sqrt {\sin ^2(e+f x)}} \]
[Out]
b*(d*sec(f*x+e))^m/f/m-a*d*hypergeom([1/2, 1/2-1/2*m],[3/2-1/2*m],cos(f*x+e)^2)*(d*sec(f*x+e))^(-1+m)*sin(f*x+
e)/f/(1-m)/(sin(f*x+e)^2)^(1/2)
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Rubi [A]
time = 0.04, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3567, 3857,
2722} \begin {gather*} \frac {b (d \sec (e+f x))^m}{f m}-\frac {a d \sin (e+f x) (d \sec (e+f x))^{m-1} \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(e+f x)\right )}{f (1-m) \sqrt {\sin ^2(e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x]),x]
[Out]
(b*(d*Sec[e + f*x])^m)/(f*m) - (a*d*Hypergeometric2F1[1/2, (1 - m)/2, (3 - m)/2, Cos[e + f*x]^2]*(d*Sec[e + f*
x])^(-1 + m)*Sin[e + f*x])/(f*(1 - m)*Sqrt[Sin[e + f*x]^2])
Rule 2722
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] && !IntegerQ[2*n]
Rule 3567
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*((d*Sec[
e + f*x])^m/(f*m)), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])
Rule 3857
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Rubi steps
\begin {align*} \int (d \sec (e+f x))^m (a+b \tan (e+f x)) \, dx &=\frac {b (d \sec (e+f x))^m}{f m}+a \int (d \sec (e+f x))^m \, dx\\ &=\frac {b (d \sec (e+f x))^m}{f m}+\left (a \left (\frac {\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{d}\right )^{-m} \, dx\\ &=\frac {b (d \sec (e+f x))^m}{f m}-\frac {a \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(e+f x)\right ) (d \sec (e+f x))^m \sin (e+f x)}{f (1-m) \sqrt {\sin ^2(e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 17.05, size = 3302, normalized size = 35.51 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x]),x]
[Out]
-((Sec[e + f*x]^(-1 - m)*(d*Sec[e + f*x])^m*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^m*(a*Sec[e + f*x]^m + b*Sec[e +
f*x]^(1 + m)*Sin[e + f*x])*Tan[(e + f*x)/2]*(-(b*AppellF1[1, m, 1 - m, 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2
]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^m*Tan[(e + f*x)/2]) - b*AppellF1[1, 1 + m, -m, 2, Tan[(e + f*x)/2]^2, -
Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^m*Tan[(e + f*x)/2] - (6*a*AppellF1[1/2, m, 1 - m, 3/2, T
an[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 + m))/(3*AppellF1[1/2, m, 1 - m, 3/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + m)*AppellF1[3/2, m, 2 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x
)/2]^2] + m*AppellF1[3/2, 1 + m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))*(a
+ b*Tan[e + f*x]))/(f*(a*Cos[e + f*x] + b*Sin[e + f*x])*(-1/2*(Sec[(e + f*x)/2]^2*(Cos[(e + f*x)/2]^2*Sec[e +
f*x])^m*(-(b*AppellF1[1, m, 1 - m, 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]
^2)^m*Tan[(e + f*x)/2]) - b*AppellF1[1, 1 + m, -m, 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*S
ec[(e + f*x)/2]^2)^m*Tan[(e + f*x)/2] - (6*a*AppellF1[1/2, m, 1 - m, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2
]^2]*(Sec[(e + f*x)/2]^2)^(-1 + m))/(3*AppellF1[1/2, m, 1 - m, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] +
2*((-1 + m)*AppellF1[3/2, m, 2 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*AppellF1[3/2, 1 + m, 1
- m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))) - (Cos[(e + f*x)/2]^2*Sec[e + f*x])^
m*Tan[(e + f*x)/2]*(-1/2*(b*AppellF1[1, m, 1 - m, 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]
^2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^m) - (b*AppellF1[1, 1 + m, -m, 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
]*Sec[(e + f*x)/2]^2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^m)/2 - b*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^m*Tan[(e + f
*x)/2]*(-1/2*((1 - m)*AppellF1[2, m, 2 - m, 3, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan
[(e + f*x)/2]) + (m*AppellF1[2, 1 + m, 1 - m, 3, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*T
an[(e + f*x)/2])/2) - b*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^m*Tan[(e + f*x)/2]*((m*AppellF1[2, 1 + m, 1 - m, 3,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/2 + ((1 + m)*AppellF1[2, 2 + m,
-m, 3, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/2) - b*m*AppellF1[1, m, 1
- m, 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1 + m)*Tan[(e + f*x)/2]*
(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e + f*x]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) - b*m*AppellF1[1, 1 +
m, -m, 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(-1 + m)*Tan[(e + f*x)/2]
*(-(Sec[(e + f*x)/2]^2*Sin[e + f*x]) + Cos[e + f*x]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) - (6*a*(-1 + m)*Appel
lF1[1/2, m, 1 - m, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 + m)*Tan[(e + f*x)/2
])/(3*AppellF1[1/2, m, 1 - m, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + m)*AppellF1[3/2, m, 2 -
m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*AppellF1[3/2, 1 + m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, -Ta
n[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) - (6*a*(Sec[(e + f*x)/2]^2)^(-1 + m)*(-1/3*((1 - m)*AppellF1[3/2, m, 2
- m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + (m*AppellF1[3/2, 1 +
m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3))/(3*AppellF1[
1/2, m, 1 - m, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + m)*AppellF1[3/2, m, 2 - m, 5/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*AppellF1[3/2, 1 + m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^
2])*Tan[(e + f*x)/2]^2) + (6*a*AppellF1[1/2, m, 1 - m, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e +
f*x)/2]^2)^(-1 + m)*(2*((-1 + m)*AppellF1[3/2, m, 2 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*Ap
pellF1[3/2, 1 + m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2] +
3*(-1/3*((1 - m)*AppellF1[3/2, m, 2 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan
[(e + f*x)/2]) + (m*AppellF1[3/2, 1 + m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]
^2*Tan[(e + f*x)/2])/3) + 2*Tan[(e + f*x)/2]^2*((-1 + m)*((-3*(2 - m)*AppellF1[5/2, m, 3 - m, 7/2, Tan[(e + f*
x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 + (3*m*AppellF1[5/2, 1 + m, 2 - m, 7/2, T
an[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5) + m*((-3*(1 - m)*AppellF1[5/2,
1 + m, 2 - m, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 + (3*(1 +
m)*AppellF1[5/2, 2 + m, 1 - m, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/
2])/5))))/(3*AppellF1[1/2, m, 1 - m, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + m)*AppellF1[3/2,
m, 2 - m, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f...
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Maple [F]
time = 0.30, size = 0, normalized size = 0.00 \[\int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^m*(a+b*tan(f*x+e)),x)
[Out]
int((d*sec(f*x+e))^m*(a+b*tan(f*x+e)),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*sec(f*x+e))^m*(a+b*tan(f*x+e)),x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e) + a)*(d*sec(f*x + e))^m, 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*sec(f*x+e))^m*(a+b*tan(f*x+e)),x, algorithm="fricas")
[Out]
integral((b*tan(f*x + e) + a)*(d*sec(f*x + e))^m, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \sec {\left (e + f x \right )}\right )^{m} \left (a + b \tan {\left (e + f x \right )}\right )\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))**m*(a+b*tan(f*x+e)),x)
[Out]
Integral((d*sec(e + f*x))**m*(a + b*tan(e + f*x)), 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*sec(f*x+e))^m*(a+b*tan(f*x+e)),x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e) + a)*(d*sec(f*x + e))^m, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right ) \,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)),x)
[Out]
int((d/cos(e + f*x))^m*(a + b*tan(e + f*x)), x)
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