3.126 \(\int \frac{(g \tan (e+f x))^p}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=108 \[ \frac{(g \tan (e+f x))^{p+1}}{a f g (p+1)}-\frac{\sec (e+f x) \cos ^2(e+f x)^{\frac{p+3}{2}} (g \tan (e+f x))^{p+2} \, _2F_1\left (\frac{p+2}{2},\frac{p+3}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{a f g^2 (p+2)} \]

[Out]

(g*Tan[e + f*x])^(1 + p)/(a*f*g*(1 + p)) - ((Cos[e + f*x]^2)^((3 + p)/2)*Hypergeometric2F1[(2 + p)/2, (3 + p)/
2, (4 + p)/2, Sin[e + f*x]^2]*Sec[e + f*x]*(g*Tan[e + f*x])^(2 + p))/(a*f*g^2*(2 + p))

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Rubi [A]  time = 0.12612, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2706, 2607, 32, 2617} \[ \frac{(g \tan (e+f x))^{p+1}}{a f g (p+1)}-\frac{\sec (e+f x) \cos ^2(e+f x)^{\frac{p+3}{2}} (g \tan (e+f x))^{p+2} \, _2F_1\left (\frac{p+2}{2},\frac{p+3}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{a f g^2 (p+2)} \]

Antiderivative was successfully verified.

[In]

Int[(g*Tan[e + f*x])^p/(a + a*Sin[e + f*x]),x]

[Out]

(g*Tan[e + f*x])^(1 + p)/(a*f*g*(1 + p)) - ((Cos[e + f*x]^2)^((3 + p)/2)*Hypergeometric2F1[(2 + p)/2, (3 + p)/
2, (4 + p)/2, Sin[e + f*x]^2]*Sec[e + f*x]*(g*Tan[e + f*x])^(2 + p))/(a*f*g^2*(2 + p))

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin{align*} \int \frac{(g \tan (e+f x))^p}{a+a \sin (e+f x)} \, dx &=\frac{\int \sec ^2(e+f x) (g \tan (e+f x))^p \, dx}{a}-\frac{\int \sec (e+f x) (g \tan (e+f x))^{1+p} \, dx}{a g}\\ &=-\frac{\cos ^2(e+f x)^{\frac{3+p}{2}} \, _2F_1\left (\frac{2+p}{2},\frac{3+p}{2};\frac{4+p}{2};\sin ^2(e+f x)\right ) \sec (e+f x) (g \tan (e+f x))^{2+p}}{a f g^2 (2+p)}+\frac{\operatorname{Subst}\left (\int (g x)^p \, dx,x,\tan (e+f x)\right )}{a f}\\ &=\frac{(g \tan (e+f x))^{1+p}}{a f g (1+p)}-\frac{\cos ^2(e+f x)^{\frac{3+p}{2}} \, _2F_1\left (\frac{2+p}{2},\frac{3+p}{2};\frac{4+p}{2};\sin ^2(e+f x)\right ) \sec (e+f x) (g \tan (e+f x))^{2+p}}{a f g^2 (2+p)}\\ \end{align*}

Mathematica [B]  time = 4.03293, size = 232, normalized size = 2.15 \[ \frac{2 \tan \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 (g \tan (e+f x))^p \left (\left (p^2+5 p+6\right ) \, _2F_1\left (\frac{p+1}{2},p+2;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-(p+1) \tan \left (\frac{1}{2} (e+f x)\right ) \left (2 (p+3) \, _2F_1\left (\frac{p+2}{2},p+2;\frac{p+4}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-(p+2) \tan \left (\frac{1}{2} (e+f x)\right ) \, _2F_1\left (p+2,\frac{p+3}{2};\frac{p+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right ) \left (\cos (e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right )\right )^p}{f (p+1) (p+2) (p+3) (a \sin (e+f x)+a)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*Tan[e + f*x])^p/(a + a*Sin[e + f*x]),x]

[Out]

(2*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^p*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*Tan[(e + f*x)/2]*((6 + 5*p + p^
2)*Hypergeometric2F1[(1 + p)/2, 2 + p, (3 + p)/2, Tan[(e + f*x)/2]^2] - (1 + p)*Tan[(e + f*x)/2]*(2*(3 + p)*Hy
pergeometric2F1[(2 + p)/2, 2 + p, (4 + p)/2, Tan[(e + f*x)/2]^2] - (2 + p)*Hypergeometric2F1[2 + p, (3 + p)/2,
(5 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]))*(g*Tan[e + f*x])^p)/(f*(1 + p)*(2 + p)*(3 + p)*(a + a*Sin[e
+ f*x]))

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Maple [F]  time = 0.178, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( g\tan \left ( fx+e \right ) \right ) ^{p}}{a+a\sin \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*tan(f*x+e))^p/(a+a*sin(f*x+e)),x)

[Out]

int((g*tan(f*x+e))^p/(a+a*sin(f*x+e)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((g*tan(f*x + e))^p/(a*sin(f*x + e) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (g \tan \left (f x + e\right )\right )^{p}}{a \sin \left (f x + e\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((g*tan(f*x + e))^p/(a*sin(f*x + e) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (g \tan{\left (e + f x \right )}\right )^{p}}{\sin{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))**p/(a+a*sin(f*x+e)),x)

[Out]

Integral((g*tan(e + f*x))**p/(sin(e + f*x) + 1), x)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((g*tan(f*x + e))^p/(a*sin(f*x + e) + a), x)