3.451 \(\int \cot ^4(e+f x) (a+b \sec ^2(e+f x))^p \, dx\)
Optimal. Leaf size=88 \[ -\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (-\frac{3}{2};1,-p;-\frac{1}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )}{3 f} \]
[Out]
-(AppellF1[-3/2, 1, -p, -1/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Cot[e + f*x]^3*(a + b + b*Tan[e
+ f*x]^2)^p)/(3*f*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)
________________________________________________________________________________________
Rubi [A] time = 0.146109, antiderivative size = 88, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used =
{4141, 1975, 511, 510} \[ -\frac{\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} F_1\left (-\frac{3}{2};1,-p;-\frac{1}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )}{3 f} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
[Out]
-(AppellF1[-3/2, 1, -p, -1/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Cot[e + f*x]^3*(a + b + b*Tan[e
+ f*x]^2)^p)/(3*f*(1 + (b*Tan[e + f*x]^2)/(a + b))^p)
Rule 4141
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])
Rule 1975
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&& !BinomialMatchQ[{u, v}, x]
Rule 511
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && !(IntegerQ[
p] || GtQ[a, 0])
Rule 510
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])
Rubi steps
\begin{align*} \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \left (1+x^2\right )\right )^p}{x^4 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^p}{x^4 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (\left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a+b}\right )^p}{x^4 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{F_1\left (-\frac{3}{2};1,-p;-\frac{1}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right ) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}}{3 f}\\ \end{align*}
Mathematica [B] time = 18.8349, size = 3033, normalized size = 34.47 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
[Out]
((a + 2*b + a*Cos[2*(e + f*x)])^p*Cot[e + f*x]^7*(Sec[e + f*x]^2)^p*(a + b*Sec[e + f*x]^2)^p*((9*(a + b)*Appel
lF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(3*(a + b)
*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/
2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a
+ b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[e + f*x]^2)/(a + b))]
- 3*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/(1 + (b*Tan[e + f*x]^2)/(a
+ b))^p))/(3*f*((2*p*(a + 2*b + a*Cos[2*(e + f*x)])^p*Cot[e + f*x]^2*(Sec[e + f*x]^2)^p*((9*(a + b)*AppellF1[
1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(3*(a + b)*App
ellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -
((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b
)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[e + f*x]^2)/(a + b))] - 3*
Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/(1 + (b*Tan[e + f*x]^2)/(a + b
))^p))/3 - (a + 2*b + a*Cos[2*(e + f*x)])^p*Cot[e + f*x]^2*Csc[e + f*x]^2*(Sec[e + f*x]^2)^p*((9*(a + b)*Appel
lF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(3*(a + b)
*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/
2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a
+ b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[e + f*x]^2)/(a + b))]
- 3*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/(1 + (b*Tan[e + f*x]^2)/(a
+ b))^p) - (2*a*p*(a + 2*b + a*Cos[2*(e + f*x)])^(-1 + p)*Cot[e + f*x]^3*(Sec[e + f*x]^2)^p*Sin[2*(e + f*x)]*
((9*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f
*x]^2)/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[
3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*T
an[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[e + f
*x]^2)/(a + b))] - 3*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/(1 + (b*T
an[e + f*x]^2)/(a + b))^p))/3 + ((a + 2*b + a*Cos[2*(e + f*x)])^p*Cot[e + f*x]^3*(Sec[e + f*x]^2)^p*((18*(a +
b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f*x])/(3*(
a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p,
1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]
^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (18*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(
a + b)), -Tan[e + f*x]^2]*Tan[e + f*x]^3)/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)),
-Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a +
b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (9*(a + b)*Sin
[e + f*x]^2*Tan[e + f*x]^2*((2*b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2
]*Sec[e + f*x]^2*Tan[e + f*x])/(3*(a + b)) - (2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[
e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/3))/(3*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)
), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a
+ b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (2*b*p*Sec[
e + f*x]^2*Tan[e + f*x]*(1 + (b*Tan[e + f*x]^2)/(a + b))^(-1 - p)*(Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[
e + f*x]^2)/(a + b))] - 3*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2))/(a
+ b) - (9*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan
[e + f*x]^2*(4*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*App
ellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2])*Sec[e + f*x]^2*Tan[e + f*x] + 3*(a + b
)*((2*b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e +
f*x])/(3*(a + b)) - (2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^
2*Tan[e + f*x])/3) + 2*Tan[e + f*x]^2*(b*p*((-6*AppellF1[5/2, 1 - p, 2, 7/2, -((b*Tan[e + f*x]^2)/(a + b)), -T
an[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/5 - (6*b*(1 - p)*AppellF1[5/2, 2 - p, 1, 7/2, -((b*Tan[e + f*x]^2)
/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*(a + b))) - (a + b)*((6*b*p*AppellF1[5/2, 1 - p, 2
, 7/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*(a + b)) - (12*AppellF1
[5/2, -p, 3, 7/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/5))))/(3*(a + b
)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*AppellF1[3/2, 1 - p, 1, 5
/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(
a + b)), -Tan[e + f*x]^2])*Tan[e + f*x]^2)^2 - (-6*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/(a +
b))]*Sec[e + f*x]^2*Tan[e + f*x] - 3*Sec[e + f*x]^2*Tan[e + f*x]*(Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e
+ f*x]^2)/(a + b))] - (1 + (b*Tan[e + f*x]^2)/(a + b))^p) - 3*Csc[e + f*x]*Sec[e + f*x]*(-Hypergeometric2F1[-3
/2, -p, -1/2, -((b*Tan[e + f*x]^2)/(a + b))] + (1 + (b*Tan[e + f*x]^2)/(a + b))^p))/(1 + (b*Tan[e + f*x]^2)/(a
+ b))^p))/3))
________________________________________________________________________________________
Maple [F] time = 0.4, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( fx+e \right ) \right ) ^{4} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
[Out]
int(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e)^2 + a)^p*cot(f*x + e)^4, x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{4}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")
[Out]
integral((b*sec(f*x + e)^2 + a)^p*cot(f*x + e)^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(cot(f*x+e)**4*(a+b*sec(f*x+e)**2)**p,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{4}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e)^2 + a)^p*cot(f*x + e)^4, x)