3.2.63 \(\int \csc ^6(e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [163]

Optimal. Leaf size=180 \[ -\frac {(10 a-b (3-2 p)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (15 a^2-b (10 a-b (3-2 p)) (1-2 p)\right ) \cot (e+f x) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a}\right ) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{15 a^2 f} \]

[Out]

-1/15*(10*a-b*(3-2*p))*cot(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1+p)/a^2/f-1/5*cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1+p)/a
/f-1/15*(15*a^2-b*(10*a-b*(3-2*p))*(1-2*p))*cot(f*x+e)*hypergeom([-1/2, -p],[1/2],-b*tan(f*x+e)^2/a)*(a+b*tan(
f*x+e)^2)^p/a^2/f/((1+b*tan(f*x+e)^2/a)^p)

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Rubi [A]
time = 0.13, antiderivative size = 176, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3744, 473, 464, 372, 371} \begin {gather*} -\frac {\left (15-\frac {b (1-2 p) (10 a-b (3-2 p))}{a^2}\right ) \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} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a}\right )}{15 f}-\frac {(10 a-b (3-2 p)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{5 a f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

-1/15*((10*a - b*(3 - 2*p))*Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(1 + p))/(a^2*f) - (Cot[e + f*x]^5*(a + b*Ta
n[e + f*x]^2)^(1 + p))/(5*a*f) - ((15 - (b*(10*a - b*(3 - 2*p))*(1 - 2*p))/a^2)*Cot[e + f*x]*Hypergeometric2F1
[-1/2, -p, 1/2, -((b*Tan[e + f*x]^2)/a)]*(a + b*Tan[e + f*x]^2)^p)/(15*f*(1 + (b*Tan[e + f*x]^2)/a)^p)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 464

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

Rule 473

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[c^2*(e*x)^(m
+ 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 3744

Int[sin[(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^(m + 1)/f), Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)
^(m/2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \csc ^6(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2 \left (a+b x^2\right )^p}{x^6} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{5 a f}+\frac {\text {Subst}\left (\int \frac {\left (10 a-b (3-2 p)+5 a x^2\right ) \left (a+b x^2\right )^p}{x^4} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {(10 a-b (3-2 p)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (-15 a^2-b (10 a-b (3-2 p)) (-3+2 (1+p))\right ) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac {(10 a-b (3-2 p)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (\left (-15 a^2-b (10 a-b (3-2 p)) (-3+2 (1+p))\right ) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a}\right )^p}{x^2} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac {(10 a-b (3-2 p)) \cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (15 a^2-b (10 a-b (3-2 p)) (1-2 p)\right ) \cot (e+f x) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a}\right ) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{15 a^2 f}\\ \end {align*}

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Mathematica [A]
time = 0.71, size = 141, normalized size = 0.78 \begin {gather*} -\frac {\cot (e+f x) \left (3 \cot ^4(e+f x) \, _2F_1\left (-\frac {5}{2},-p;-\frac {3}{2};-\frac {b \tan ^2(e+f x)}{a}\right )+10 \cot ^2(e+f x) \, _2F_1\left (-\frac {3}{2},-p;-\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a}\right )+15 \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \tan ^2(e+f x)}{a}\right )\right ) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

-1/15*(Cot[e + f*x]*(3*Cot[e + f*x]^4*Hypergeometric2F1[-5/2, -p, -3/2, -((b*Tan[e + f*x]^2)/a)] + 10*Cot[e +
f*x]^2*Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[e + f*x]^2)/a)] + 15*Hypergeometric2F1[-1/2, -p, 1/2, -((b*T
an[e + f*x]^2)/a)])*(a + b*Tan[e + f*x]^2)^p)/(f*(1 + (b*Tan[e + f*x]^2)/a)^p)

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \left (\csc ^{6}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(csc(f*x+e)^6*(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(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^6, 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(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^6, x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**6*(a+b*tan(f*x+e)**2)**p,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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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(csc(f*x+e)^6*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p}{{\sin \left (e+f\,x\right )}^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x)^2)^p/sin(e + f*x)^6,x)

[Out]

int((a + b*tan(e + f*x)^2)^p/sin(e + f*x)^6, x)

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