3.5.44 \(\int (a+b \sec ^2(e+f x))^p \tan (e+f x) \, dx\) [444]

Optimal. Leaf size=54 \[ -\frac {\, _2F_1\left (1,1+p;2+p;1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4224, 272, 67} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sec[e + f*x]^2)/a]*(a + b*Sec[e + f*x]^2)^(1 + p))/(a*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 272

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

Rule 4224

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])

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 54, normalized size = 1.00 \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p} \tan \left (f x +e \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \mathrm {tan}\left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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