\(\int (a+b \tan (c+d x))^n \, dx\) [711]

Optimal result

Integrand size = 12, antiderivative size = 167 \[ \int (a+b \tan (c+d x))^n \, dx=\frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right ) d (1+n)}-\frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right ) d (1+n)} \]

[Out]

1/2*b*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a-(-b^2)^(1/2)))*(a+b*tan(d*x+c))^(1+n)/d/(1+n)/(a-(-b^2)^(1/
2))/(-b^2)^(1/2)-1/2*b*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)))*(a+b*tan(d*x+c))^(1+n)/d/(1
+n)/(-b^2)^(1/2)/(a+(-b^2)^(1/2))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3566, 726, 70} \[ \int (a+b \tan (c+d x))^n \, dx=\frac {b (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a-\sqrt {-b^2}\right )}-\frac {b (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{2 \sqrt {-b^2} d (n+1) \left (a+\sqrt {-b^2}\right )} \]

[In]

Int[(a + b*Tan[c + d*x])^n,x]

[Out]

(b*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*
Sqrt[-b^2]*(a - Sqrt[-b^2])*d*(1 + n)) - (b*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[
-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*Sqrt[-b^2]*(a + Sqrt[-b^2])*d*(1 + n))

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 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 3566

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {(a+x)^n}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {b \text {Subst}\left (\int \left (\frac {\sqrt {-b^2} (a+x)^n}{2 b^2 \left (\sqrt {-b^2}-x\right )}+\frac {\sqrt {-b^2} (a+x)^n}{2 b^2 \left (\sqrt {-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{d} \\ & = -\frac {b \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{2 \sqrt {-b^2} d}-\frac {b \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{2 \sqrt {-b^2} d} \\ & = \frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right ) d (1+n)}-\frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right ) d (1+n)} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.71 \[ \int (a+b \tan (c+d x))^n \, dx=\frac {\left ((a+i b) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-i b}\right )-(a-i b) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))^{1+n}}{2 (a+i b) (i a+b) d (1+n)} \]

[In]

Integrate[(a + b*Tan[c + d*x])^n,x]

[Out]

(((a + I*b)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - I*b)] - (a - I*b)*Hypergeometric2F1[1
, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[c + d*x])^(1 + n))/(2*(a + I*b)*(I*a + b)*d*(1 + n
))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b*tan(d*x + c) + a)^n, x)

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*tan(c + d*x))**n, x)

Maxima [F]

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

[In]

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

[Out]

integrate((b*tan(d*x + c) + a)^n, x)

Giac [F]

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

[In]

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

[Out]

integrate((b*tan(d*x + c) + a)^n, x)

Mupad [F(-1)]

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

[In]

int((a + b*tan(c + d*x))^n,x)

[Out]

int((a + b*tan(c + d*x))^n, x)