\(\int (a+b \sec (c+d x))^n \tan ^3(c+d x) \, dx\) [354]

Optimal result

Integrand size = 21, antiderivative size = 102 \[ \int (a+b \sec (c+d x))^n \tan ^3(c+d x) \, dx=-\frac {a (a+b \sec (c+d x))^{1+n}}{b^2 d (1+n)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}+\frac {(a+b \sec (c+d x))^{2+n}}{b^2 d (2+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3970, 966, 81, 67} \[ \int (a+b \sec (c+d x))^n \tan ^3(c+d x) \, dx=-\frac {a (a+b \sec (c+d x))^{n+1}}{b^2 d (n+1)}+\frac {(a+b \sec (c+d x))^{n+2}}{b^2 d (n+2)}+\frac {(a+b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \]

[In]

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

[Out]

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

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 966

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[c^p*(d
+ e*x)^(m + 2*p)*((f + g*x)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Dist[1/(g*e^(2*p)*(m + n + 2*p + 1)),
Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c
^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0]
&& NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 3970

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[(b^2 - x^2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int (a+b \sec (c+d x))^n \tan ^3(c+d x) \, dx=\frac {(a+b \sec (c+d x))^{1+n} \left (b^2 (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {b \sec (c+d x)}{a}\right )+a (-a+b (1+n) \sec (c+d x))\right )}{a b^2 d (1+n) (2+n)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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