\(\int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx\) [651]

Optimal result

Integrand size = 21, antiderivative size = 159 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) \sec (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (1+n) \sqrt {1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3593, 774, 138} \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\sec (c+d x) (a+b \tan (c+d x))^{n+1} \operatorname {AppellF1}\left (n+1,-\frac {1}{2},-\frac {1}{2},n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{b d (n+1) \sqrt {1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}}} \]

[In]

Int[Sec[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]

[Out]

(AppellF1[1 + n, -1/2, -1/2, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]), (a + b*Tan[c + d*x])/(a + Sqrt[-b^2
])]*Sec[c + d*x]*(a + b*Tan[c + d*x])^(1 + n))/(b*d*(1 + n)*Sqrt[1 - (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])]*Sq
rt[1 - (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])])

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 774

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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sec (c+d x) \text {Subst}\left (\int (a+x)^n \sqrt {1+\frac {x^2}{b^2}} \, dx,x,b \tan (c+d x)\right )}{b d \sqrt {\sec ^2(c+d x)}} \\ & = \frac {\sec (c+d x) \text {Subst}\left (\int x^n \sqrt {1-\frac {x}{a-\sqrt {-b^2}}} \sqrt {1-\frac {x}{a+\sqrt {-b^2}}} \, dx,x,a+b \tan (c+d x)\right )}{b d \sqrt {1-\frac {a+b \tan (c+d x)}{a-\frac {b^2}{\sqrt {-b^2}}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\frac {b^2}{\sqrt {-b^2}}}}} \\ & = \frac {\operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) \sec (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (1+n) \sqrt {1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}} \sqrt {1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 14.54 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.92 \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {2 (a-i b) (a+i b) (2+n) \operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right ) \sec (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (1+n) \left (2 \left (a^2+b^2\right ) (2+n) \operatorname {AppellF1}\left (1+n,-\frac {1}{2},-\frac {1}{2},2+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )-\left ((a-i b) \operatorname {AppellF1}\left (2+n,-\frac {1}{2},\frac {1}{2},3+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )+(a+i b) \operatorname {AppellF1}\left (2+n,\frac {1}{2},-\frac {1}{2},3+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))\right )} \]

[In]

Integrate[Sec[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]

[Out]

(2*(a - I*b)*(a + I*b)*(2 + n)*AppellF1[1 + n, -1/2, -1/2, 2 + n, (a + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c
+ d*x])/(a + I*b)]*Sec[c + d*x]*(a + b*Tan[c + d*x])^(1 + n))/(b*d*(1 + n)*(2*(a^2 + b^2)*(2 + n)*AppellF1[1
+ n, -1/2, -1/2, 2 + n, (a + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)] - ((a - I*b)*AppellF1[
2 + n, -1/2, 1/2, 3 + n, (a + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)] + (a + I*b)*AppellF1[
2 + n, 1/2, -1/2, 3 + n, (a + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[c + d*x])
))

Maple [F]

\[\int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

[In]

int(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x)

[Out]

int(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x)

Fricas [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \sec ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**3*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*sec(c + d*x)**3, x)

Maxima [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)

Giac [F]

\[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+b \tan (c+d x))^n \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^3} \,d x \]

[In]

int((a + b*tan(c + d*x))^n/cos(c + d*x)^3,x)

[Out]

int((a + b*tan(c + d*x))^n/cos(c + d*x)^3, x)