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

Optimal. Leaf size=140 \[ \frac{b \left (6 a^2+b^2 \left (n^2-3 n+2\right )\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )}{6 a^4 d (n+1)}+\frac{b (2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{6 a^2 d}-\frac{\cot ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{3 a d} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.12994, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3516, 950, 78, 65} \[ \frac{b \left (6 a^2+b^2 \left (n^2-3 n+2\right )\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )}{6 a^4 d (n+1)}+\frac{b (2-n) \cot ^2(c+d x) (a+b \tan (c+d x))^{n+1}}{6 a^2 d}-\frac{\cot ^3(c+d x) (a+b \tan (c+d x))^{n+1}}{3 a d} \]

Antiderivative was successfully verified.

[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 3516

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]

Rule 950

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 78

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] || LtQ
[p, n]))))

Rule 65

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

Rubi steps

\begin{align*} \int \csc ^4(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac{b \operatorname{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 \operatorname{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 ) \operatorname{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 ) \, _2F_1\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]  time = 1.24389, size = 78, normalized size = 0.56 \[ \frac{b (a+b \tan (c+d x))^{n+1} \left (a^2 \, _2F_1\left (2,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )+b^2 \, _2F_1\left (4,n+1;n+2;\frac{b \tan (c+d x)}{a}+1\right )\right )}{a^4 d (n+1)} \]

Antiderivative was successfully verified.

[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]  time = 0.229, size = 0, normalized size = 0. \begin{align*} \int \left ( \csc \left ( dx+c \right ) \right ) ^{4} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)