\(\int \cot ^3(c+d x) (a+b \sin ^2(c+d x))^p \, dx\) [547]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=-\frac {\csc ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 a d}+\frac {(a-b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^2(c+d x)}{a}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 a^2 d (1+p)} \]

[Out]

-1/2*csc(d*x+c)^2*(a+b*sin(d*x+c)^2)^(p+1)/a/d+1/2*(-b*p+a)*hypergeom([1, p+1],[2+p],1+b*sin(d*x+c)^2/a)*(a+b*
sin(d*x+c)^2)^(p+1)/a^2/d/(p+1)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3273, 79, 67} \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\frac {(a-b p) \left (a+b \sin ^2(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^2(c+d x)}{a}+1\right )}{2 a^2 d (p+1)}-\frac {\csc ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{p+1}}{2 a d} \]

[In]

Int[Cot[c + d*x]^3*(a + b*Sin[c + d*x]^2)^p,x]

[Out]

-1/2*(Csc[c + d*x]^2*(a + b*Sin[c + d*x]^2)^(1 + p))/(a*d) + ((a - b*p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 +
(b*Sin[c + d*x]^2)/a]*(a + b*Sin[c + d*x]^2)^(1 + p))/(2*a^2*d*(1 + p))

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 3273

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m
+ 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

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

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.77 \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=-\frac {\left (a \csc ^2(c+d x)+\frac {(-a+b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^2(c+d x)}{a}\right )}{1+p}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 a^2 d} \]

[In]

Integrate[Cot[c + d*x]^3*(a + b*Sin[c + d*x]^2)^p,x]

[Out]

-1/2*((a*Csc[c + d*x]^2 + ((-a + b*p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^2)/a])/(1 + p))*(
a + b*Sin[c + d*x]^2)^(1 + p))/(a^2*d)

Maple [F]

\[\int \left (\cot ^{3}\left (d x +c \right )\right ) {\left (a +\left (\sin ^{2}\left (d x +c \right )\right ) b \right )}^{p}d x\]

[In]

int(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x)

[Out]

int(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x)

Fricas [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x, algorithm="fricas")

[Out]

integral((-b*cos(d*x + c)^2 + a + b)^p*cot(d*x + c)^3, x)

Sympy [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**3*(a+b*sin(d*x+c)**2)**p,x)

[Out]

Timed out

Maxima [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*cot(d*x + c)^3, x)

Giac [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*cot(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (b\,{\sin \left (c+d\,x\right )}^2+a\right )}^p \,d x \]

[In]

int(cot(c + d*x)^3*(a + b*sin(c + d*x)^2)^p,x)

[Out]

int(cot(c + d*x)^3*(a + b*sin(c + d*x)^2)^p, x)