3.442 \(\int (a+b \sec ^2(e+f x))^p \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=122 \[ -\frac{\left (a+b \sec ^2(e+f x)\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{2 b^2 f (p+1)}+\frac{\left (a+b \sec ^2(e+f x)\right )^{p+2}}{2 b^2 f (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.14756, antiderivative size = 122, normalized size of antiderivative = 1., 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 = {4139, 446, 88, 65} \[ -\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{2 b^2 f (p+1)}+\frac{\left (a+b \sec ^2(e+f x)\right )^{p+2}}{2 b^2 f (p+2)}-\frac{\left (a+b \sec ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rule 446

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 88

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 65

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

Rubi steps

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

Mathematica [A]  time = 0.599999, size = 94, normalized size = 0.77 \[ -\frac{\left (a+b \sec ^2(e+f x)\right )^{p+1} \left (b^2 (p+2) \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \sec ^2(e+f x)}{a}+1\right )+a \left (a-b (p+1) \sec ^2(e+f x)+2 b (p+2)\right )\right )}{2 a b^2 f (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.487, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p} \left ( \tan \left ( fx+e \right ) \right ) ^{5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

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

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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} \tan \left (f x + e\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*tan(f*x+e)^5,x, algorithm="giac")

[Out]

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