Integrand size = 21, antiderivative size = 140 \[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b (2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{1+n}}{6 a^2 d}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{1+n}}{3 a d}+\frac {b \left (6 a^2+b^2 \left (2-3 n+n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right ) (a+b \tan (c+d x))^{1+n}}{6 a^4 d (1+n)} \]
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)
Time = 7.97 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.56 \[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b \left (a^2 \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right )+b^2 \operatorname {Hypergeometric2F1}\left (4,1+n,2+n,1+\frac {b \tan (c+d x)}{a}\right )\right ) (a+b \tan (c+d x))^{1+n}}{a^4 d (1+n)} \]
(b*(a^2*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a] + b^2*H ypergeometric2F1[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))
Time = 0.30 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3999, 520, 87, 75}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+b \tan (c+d x))^n}{\sin (c+d x)^4}dx\) |
\(\Big \downarrow \) 3999 |
\(\displaystyle \frac {b \int \frac {\cot ^4(c+d x) (a+b \tan (c+d x))^n \left (\tan ^2(c+d x) b^2+b^2\right )}{b^4}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 520 |
\(\displaystyle \frac {b \left (-\frac {\int \frac {\cot ^3(c+d x) (a+b \tan (c+d x))^n \left (b^2 (2-n)-3 a b \tan (c+d x)\right )}{b^3}d(b \tan (c+d x))}{3 a}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{3 a b}\right )}{d}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {b \left (-\frac {-\frac {\left (6 a^2+b^2 (1-n) (2-n)\right ) \int \frac {\cot ^2(c+d x) (a+b \tan (c+d x))^n}{b^2}d(b \tan (c+d x))}{2 a}-\frac {(2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{2 a}}{3 a}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{3 a b}\right )}{d}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {b \left (-\frac {-\frac {\left (6 a^2+b^2 (1-n) (2-n)\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {b \tan (c+d x)}{a}+1\right )}{2 a^3 (n+1)}-\frac {(2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{2 a}}{3 a}-\frac {\cot ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{3 a b}\right )}{d}\) |
(b*(-1/3*(Cot[c + d*x]^3*(a + b*Tan[c + d*x])^(1 + n))/(a*b) - (-1/2*((2 - n)*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n))/a - ((6*a^2 + b^2*(1 - n) *(2 - n))*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Tan[c + d*x])/a]*(a + b*Tan[c + d*x])^(1 + n))/(2*a^3*(1 + n)))/(3*a)))/d
3.1.87.3.1 Defintions of rubi rules used
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] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(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] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, e*x, x], R = Pol ynomialRemainder[(a + b*x^2)^p, e*x, x]}, Simp[R*(e*x)^(m + 1)*((c + d*x)^( n + 1)/((m + 1)*(e*c))), x] + Simp[1/((m + 1)*(e*c)) Int[(e*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(e*c)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[m, -1] && !IntegerQ[n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[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]
\[\int \left (\csc ^{4}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]
\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \csc ^{4}{\left (c + d x \right )}\, dx \]
\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
\[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int { {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4} \,d x } \]
Timed out. \[ \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n}{{\sin \left (c+d\,x\right )}^4} \,d x \]