3.371 \(\int \cot ^2(e+f x) (a+b \tan ^2(e+f x))^p \, dx\)
Optimal. Leaf size=79 \[ -\frac{\cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (-\frac{1}{2};1,-p;\frac{1}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f} \]
[Out]
-((AppellF1[-1/2, 1, -p, 1/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Cot[e + f*x]*(a + b*Tan[e + f*x]^2)^p)
/(f*(1 + (b*Tan[e + f*x]^2)/a)^p))
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Rubi [A] time = 0.0984764, antiderivative size = 79, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used =
{3670, 511, 510} \[ -\frac{\cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (-\frac{1}{2};1,-p;\frac{1}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f} \]
Antiderivative was successfully verified.
[In]
Int[Cot[e + f*x]^2*(a + b*Tan[e + f*x]^2)^p,x]
[Out]
-((AppellF1[-1/2, 1, -p, 1/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Cot[e + f*x]*(a + b*Tan[e + f*x]^2)^p)
/(f*(1 + (b*Tan[e + f*x]^2)/a)^p))
Rule 3670
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))
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 ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{x^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\left (\left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a}\right )^p}{x^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{F_1\left (-\frac{1}{2};1,-p;\frac{1}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right ) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a}\right )^{-p}}{f}\\ \end{align*}
Mathematica [B] time = 14.5048, size = 1989, normalized size = 25.18 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[e + f*x]^2*(a + b*Tan[e + f*x]^2)^p,x]
[Out]
(Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(2*p)*(-(Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/a)]/(1 +
(b*Tan[e + f*x]^2)/a)^p) + (3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sin[e + f
*x]^2)/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 -
p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -
Tan[e + f*x]^2])*Tan[e + f*x]^2)))/(f*(2*b*p*Sec[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(-1 + p)*(-(Hypergeometric2
F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/a)]/(1 + (b*Tan[e + f*x]^2)/a)^p) + (3*a*AppellF1[1/2, -p, 1, 3/2, -((b
*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sin[e + f*x]^2)/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a),
-Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*Appel
lF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2)) - Csc[e + f*x]^2*(a + b*Tan[e
+ f*x]^2)^p*(-(Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/a)]/(1 + (b*Tan[e + f*x]^2)/a)^p) + (3*a
*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sin[e + f*x]^2)/(-3*a*AppellF1[1/2, -p, 1
, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/
a), -Tan[e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2)
) + Cot[e + f*x]*(a + b*Tan[e + f*x]^2)^p*((2*b*p*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/a)]*Se
c[e + f*x]^2*Tan[e + f*x]*(1 + (b*Tan[e + f*x]^2)/a)^(-1 - p))/a + (6*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e +
f*x]^2)/a), -Tan[e + f*x]^2]*Cos[e + f*x]*Sin[e + f*x])/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/
a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*Ap
pellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2) + (3*a*Sin[e + f*x]^2*((2*b
*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(3*a) -
(2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/3))/(-3*a
*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -
((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]
^2])*Tan[e + f*x]^2) - (Csc[e + f*x]*Sec[e + f*x]*(Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/a)] -
(1 + (b*Tan[e + f*x]^2)/a)^p))/(1 + (b*Tan[e + f*x]^2)/a)^p - (3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x
]^2)/a), -Tan[e + f*x]^2]*Sin[e + f*x]^2*(4*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[
e + f*x]^2]) + a*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Sec[e + f*x]^2*Tan[e + f
*x] - 3*a*((2*b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e
+ f*x])/(3*a) - (2*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -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), -Tan[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), -Tan[e +
f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*a))) + a*((6*b*p*AppellF1[5/2, 1 - p, 2, 7/2, -((b*Tan[e + f*x]^2)/a),
-Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/(5*a) - (12*AppellF1[5/2, -p, 3, 7/2, -((b*Tan[e + f*x]^2)/a),
-Tan[e + f*x]^2]*Sec[e + f*x]^2*Tan[e + f*x])/5))))/(-3*a*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/a), -
Tan[e + f*x]^2] + 2*(-(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2]) + a*AppellF
1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/a), -Tan[e + f*x]^2])*Tan[e + f*x]^2)^2)))
________________________________________________________________________________________
Maple [F] time = 0.281, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \tan \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)^2*(a+b*tan(f*x+e)^2)^p,x)
[Out]
int(cot(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e)^2 + a)^p*cot(f*x + e)^2, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{2}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")
[Out]
integral((b*tan(f*x + e)^2 + a)^p*cot(f*x + e)^2, 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)**2*(a+b*tan(f*x+e)**2)**p,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(f*x+e)^2*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e)^2 + a)^p*cot(f*x + e)^2, x)