\(\int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx\) [87]

Optimal result

Integrand size = 21, antiderivative size = 140 \[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b (2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{6 a^2 d}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{3 a d}+\frac {b \left (6 a^2+b^2 \left (2-3 n+n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{6 a^4 d (1+n)} \]

[Out]

1/6*b*(2-n)*cot(d*x+c)^2*(a+b*tan(d*x+c))^(1+n)/a^2/d-1/3*cot(d*x+c)^3*(a+b*tan(d*x+c))^(1+n)/a/d+1/6*b*(6*a^2
+b^2*(n^2-3*n+2))*hypergeom([2, 1+n],[2+n],1+b*tan(d*x+c)/a)*(a+b*tan(d*x+c))^(1+n)/a^4/d/(1+n)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 140, 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 = {3597, 964, 79, 67} \[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b (2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{6 a^2 d}+\frac {b \left (6 a^2+b^2 \left (n^2-3 n+2\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {b \tan (c+d x)}{a}+1\right )}{6 a^4 d (n+1)}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{3 a d} \]

[In]

Int[Csc[c + d*x]^4*(a + b*Tan[c + d*x])^n,x]

[Out]

(b*(2 - n)*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n))/(6*a^2*d) - (Cot[c + d*x]^3*(a + b*Tan[c + d*x])^(1 +
n))/(3*a*d) + (b*(6*a^2 + b^2*(2 - 3*n + n^2))*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a]*(a +
b*Tan[c + d*x])^(1 + n))/(6*a^4*d*(1 + 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 79

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

Rule 964

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = P
olynomialQuotient[(a + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + c*x^2)^p, d + e*x, x]}, Simp[R*(d +
e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g))), x] + Dist[1/((m + 1)*(e*f - d*g)), Int[(d + e*x)^(m +
1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), 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] && LtQ[m, -1]

Rule 3597

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {(a+x)^n \left (b^2+x^2\right )}{x^4} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = -\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{3 a d}-\frac {b \text {Subst}\left (\int \frac {(a+x)^n \left (b^2 (2-n)-3 a x\right )}{x^3} \, dx,x,b \tan (c+d x)\right )}{3 a d} \\ & = \frac {b (2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{6 a^2 d}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{3 a d}-\frac {\left (b \left (-6 a^2+b^2 (2-n) (-1+n)\right )\right ) \text {Subst}\left (\int \frac {(a+x)^n}{x^2} \, dx,x,b \tan (c+d x)\right )}{6 a^2 d} \\ & = \frac {b (2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{6 a^2 d}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{3 a d}+\frac {b \left (6 a^2+b^2 (1-n) (2-n)\right ) \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{6 a^4 d (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.97 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.56 \[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b \left (a^2 \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right )+b^2 \operatorname {Hypergeometric2F1}\left (4,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right )\right ) (a+b \tan (c+d x))^{1+n}}{a^4 d (1+n)} \]

[In]

Integrate[Csc[c + d*x]^4*(a + b*Tan[c + d*x])^n,x]

[Out]

(b*(a^2*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a] + b^2*Hypergeometric2F1[4, 1 + n, 2 + n, 1
+ (b*Tan[c + d*x])/a])*(a + b*Tan[c + d*x])^(1 + n))/(a^4*d*(1 + n))

Maple [F]

\[\int \left (\csc ^{4}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

[In]

int(csc(d*x+c)^4*(a+b*tan(d*x+c))^n,x)

[Out]

int(csc(d*x+c)^4*(a+b*tan(d*x+c))^n,x)

Fricas [F]

\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(csc(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*tan(d*x + c) + a)^n*csc(d*x + c)^4, x)

Sympy [F]

\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \csc ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(csc(d*x+c)**4*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*csc(c + d*x)**4, x)

Maxima [F]

\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(csc(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^n*csc(d*x + c)^4, x)

Giac [F]

\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(csc(d*x+c)^4*(a+b*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^n*csc(d*x + c)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{{\sin \left (c+d\,x\right )}^4} \,d x \]

[In]

int((a + b*tan(c + d*x))^n/sin(c + d*x)^4,x)

[Out]

int((a + b*tan(c + d*x))^n/sin(c + d*x)^4, x)