3.4.61 \(\int \tan ^5(e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [361]

Optimal. Leaf size=129 \[ -\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}-\frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \tan ^2(e+f x)}{a-b}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 (a-b) f (1+p)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3751, 457, 90, 70} \begin {gather*} -\frac {(a+b) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{2 b^2 f (p+1)}+\frac {\left (a+b \tan ^2(e+f x)\right )^{p+2}}{2 b^2 f (p+2)}-\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a + b)*(a + b*Tan[e + f*x]^2)^(1 + p))/(b^2*f*(1 + p)) - (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)) + (a + b*Tan[e + f*x]^2)^(2 + p)/(
2*b^2*f*(2 + p))

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 90

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

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

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Mathematica [A]
time = 0.59, size = 106, normalized size = 0.82 \begin {gather*} \frac {\left (a+b \tan ^2(e+f x)\right )^{1+p} \left (b^2 (2+p) \, _2F_1\left (1,1+p;2+p;\frac {a+b \tan ^2(e+f x)}{a-b}\right )+(a-b) \left (a+b (2+p)-b (1+p) \tan ^2(e+f x)\right )\right )}{2 b^2 (-a+b) f (1+p) (2+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(tan(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(tan(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*tan(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(tan(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*tan(f*x + e)^5, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{p} \tan ^{5}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(tan(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*tan(f*x + e)^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\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(tan(e + f*x)^5*(a + b*tan(e + f*x)^2)^p,x)

[Out]

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

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