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

Optimal. Leaf size=181 \[ \frac {b (d \sec (e+f x))^m \left (\frac {a+b \tan (e+f x)}{\sqrt {-b^2}-a}+1\right )^{-m/2} \left (1-\frac {a+b \tan (e+f x)}{a+\sqrt {-b^2}}\right )^{-m/2} (a+b \tan (e+f x))^{n+1} F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a+\sqrt {-b^2}},\frac {a+b \tan (e+f x)}{a-\sqrt {-b^2}}\right )}{f (n+1) \left (a^2+b^2\right )} \]

[Out]

b*AppellF1(1+n,1-1/2*m,1-1/2*m,2+n,(a+b*tan(f*x+e))/(a-(-b^2)^(1/2)),(a+b*tan(f*x+e))/(a+(-b^2)^(1/2)))*(d*sec
(f*x+e))^m*(a+b*tan(f*x+e))^(1+n)/(a^2+b^2)/f/(1+n)/((1+(a+b*tan(f*x+e))/(-a+(-b^2)^(1/2)))^(1/2*m))/((1+(-a-b
*tan(f*x+e))/(a+(-b^2)^(1/2)))^(1/2*m))

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Rubi [A]  time = 0.17, antiderivative size = 187, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3512, 760, 133} \[ \frac {\cos ^2(e+f x) (d \sec (e+f x))^m \left (1-\frac {a+b \tan (e+f x)}{a-\sqrt {-b^2}}\right )^{1-\frac {m}{2}} \left (1-\frac {a+b \tan (e+f x)}{a+\sqrt {-b^2}}\right )^{1-\frac {m}{2}} (a+b \tan (e+f x))^{n+1} F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a-\sqrt {-b^2}},\frac {a+b \tan (e+f x)}{a+\sqrt {-b^2}}\right )}{b f (n+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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])

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

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Mathematica [C]  time = 6.55, size = 699, normalized size = 3.86 \[ \frac {2 (d \sec (e+f x))^m (a+b \tan (e+f x))^{n+1} F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )}{f \left (2 n (b-a \tan (e+f x)) F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )+2 (m+n) \tan (e+f x) (a+b \tan (e+f x)) F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )+2 b \sec ^2(e+f x) F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )-\frac {b (m-2) (n+1) \sec ^2(e+f x) (a+b \tan (e+f x)) \left ((a-i b) F_1\left (n+2;1-\frac {m}{2},2-\frac {m}{2};n+3;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )+(a+i b) F_1\left (n+2;2-\frac {m}{2},1-\frac {m}{2};n+3;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )\right )}{(n+2) (a-i b) (a+i b)}-\frac {m \sec ^2(e+f x) (a+b \tan (e+f x)) F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )}{\tan (e+f x)-i}-\frac {m \sec ^2(e+f x) (a+b \tan (e+f x)) F_1\left (n+1;1-\frac {m}{2},1-\frac {m}{2};n+2;\frac {a+b \tan (e+f x)}{a-i b},\frac {a+b \tan (e+f x)}{a+i b}\right )}{\tan (e+f x)+i}\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[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
*Sec[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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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (d \sec \left (f x + e\right )\right )^{m} {\left (b \tan \left (f x + e\right ) + a\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{m} {\left (b \tan \left (f x + e\right ) + a\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.93, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{m} {\left (b \tan \left (f x + e\right ) + a\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{m} \left (a + b \tan {\left (e + f x \right )}\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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