3.4.66 \(\int \cot ^5(e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [366]

Optimal. Leaf size=217 \[ \frac {(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac {\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}+\frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \tan ^2(e+f x)}{a-b}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 (a-b) f (1+p)}-\frac {\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \, _2F_1\left (1,1+p;2+p;1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^3 f (1+p)} \]

[Out]

1/4*(-b*p+2*a+b)*cot(f*x+e)^2*(a+b*tan(f*x+e)^2)^(1+p)/a^2/f-1/4*cot(f*x+e)^4*(a+b*tan(f*x+e)^2)^(1+p)/a/f+1/2
*hypergeom([1, 1+p],[2+p],(a+b*tan(f*x+e)^2)/(a-b))*(a+b*tan(f*x+e)^2)^(1+p)/(a-b)/f/(1+p)-1/4*(2*a^2-2*a*b*p-
b^2*(1-p)*p)*hypergeom([1, 1+p],[2+p],1+b*tan(f*x+e)^2/a)*(a+b*tan(f*x+e)^2)^(1+p)/a^3/f/(1+p)

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Rubi [A]
time = 0.17, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3751, 457, 105, 156, 162, 67, 70} \begin {gather*} \frac {(2 a-b p+b) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a^2 f}-\frac {\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)}{a}+1\right )}{4 a^3 f (p+1)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a}{a-b}\right )}{2 f (p+1) (a-b)}-\frac {\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

((2*a + b - b*p)*Cot[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(1 + p))/(4*a^2*f) - (Cot[e + f*x]^4*(a + b*Tan[e + f*x
]^2)^(1 + p))/(4*a*f) + (Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Tan[e + f*x]^2)/(a - b)]*(a + b*Tan[e + f*x
]^2)^(1 + p))/(2*(a - b)*f*(1 + p)) - ((2*a^2 - 2*a*b*p - b^2*(1 - p)*p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1
+ (b*Tan[e + f*x]^2)/a]*(a + b*Tan[e + f*x]^2)^(1 + p))/(4*a^3*f*(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 70

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

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3751

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]))

Rubi steps

\begin {align*} \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^5 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {(a+b x)^p}{x^3 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac {\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}-\frac {\text {Subst}\left (\int \frac {(a+b x)^p (2 a+b-b p+b (1-p) x)}{x^2 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{4 a f}\\ &=\frac {(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac {\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}+\frac {\text {Subst}\left (\int \frac {(a+b x)^p \left (2 a^2-2 a b p-b^2 (1-p) p-b p (2 a+b-b p) x\right )}{x (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac {(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac {\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}-\frac {\text {Subst}\left (\int \frac {(a+b x)^p}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}+\frac {\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac {(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac {\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}+\frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \tan ^2(e+f x)}{a-b}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 (a-b) f (1+p)}-\frac {\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \, _2F_1\left (1,1+p;2+p;1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^3 f (1+p)}\\ \end {align*}

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Mathematica [A]
time = 1.73, size = 172, normalized size = 0.79 \begin {gather*} -\frac {\left (b+a \cot ^2(e+f x)\right ) \left (-2 a^3 \, _2F_1\left (1,1+p;2+p;\frac {a+b \tan ^2(e+f x)}{a-b}\right )+(a-b) \left (a (1+p) \cot ^2(e+f x) \left (-2 a+b (-1+p)+a \cot ^2(e+f x)\right )+\left (2 a^2-2 a b p+b^2 (-1+p) p\right ) \, _2F_1\left (1,1+p;2+p;1+\frac {b \tan ^2(e+f x)}{a}\right )\right )\right ) \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^p}{4 a^3 (a-b) f (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

-1/4*((b + a*Cot[e + f*x]^2)*(-2*a^3*Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Tan[e + f*x]^2)/(a - b)] + (a -
b)*(a*(1 + p)*Cot[e + f*x]^2*(-2*a + b*(-1 + p) + a*Cot[e + f*x]^2) + (2*a^2 - 2*a*b*p + b^2*(-1 + p)*p)*Hype
rgeometric2F1[1, 1 + p, 2 + p, 1 + (b*Tan[e + f*x]^2)/a]))*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^p)/(a^3*(a -
b)*f*(1 + p))

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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int \left (\cot ^{5}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^p,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(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)^5, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(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)^5, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5*(a+b*tan(f*x+e)**2)**p,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(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)^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cot}\left (e+f\,x\right )}^5\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^5*(a + b*tan(e + f*x)^2)^p,x)

[Out]

int(cot(e + f*x)^5*(a + b*tan(e + f*x)^2)^p, x)

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