3.127 \(\int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=138 \[ \frac {2 (g \tan (e+f x))^{p+3}}{a^2 f g^3 (p+3)}-\frac {2 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac {p+5}{2}} (g \tan (e+f x))^{p+2} \, _2F_1\left (\frac {p+2}{2},\frac {p+5}{2};\frac {p+4}{2};\sin ^2(e+f x)\right )}{a^2 f g^2 (p+2)}+\frac {(g \tan (e+f x))^{p+1}}{a^2 f g (p+1)} \]

[Out]

(g*tan(f*x+e))^(1+p)/a^2/f/g/(1+p)-2*(cos(f*x+e)^2)^(5/2+1/2*p)*hypergeom([1+1/2*p, 5/2+1/2*p],[2+1/2*p],sin(f
*x+e)^2)*sec(f*x+e)^3*(g*tan(f*x+e))^(2+p)/a^2/f/g^2/(2+p)+2*(g*tan(f*x+e))^(3+p)/a^2/f/g^3/(3+p)

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Rubi [A]  time = 0.27, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2711, 2607, 14, 16, 2617, 32} \[ -\frac {2 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac {p+5}{2}} (g \tan (e+f x))^{p+2} \, _2F_1\left (\frac {p+2}{2},\frac {p+5}{2};\frac {p+4}{2};\sin ^2(e+f x)\right )}{a^2 f g^2 (p+2)}+\frac {2 (g \tan (e+f x))^{p+3}}{a^2 f g^3 (p+3)}+\frac {(g \tan (e+f x))^{p+1}}{a^2 f g (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(g*Tan[e + f*x])^p/(a + a*Sin[e + f*x])^2,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 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 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]

Rule 2711

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

Rubi steps

\begin {align*} \int \frac {(g \tan (e+f x))^p}{(a+a \sin (e+f x))^2} \, dx &=\frac {\int \left (a^2 \sec ^4(e+f x) (g \tan (e+f x))^p-2 a^2 \sec ^3(e+f x) \tan (e+f x) (g \tan (e+f x))^p+a^2 \sec ^2(e+f x) \tan ^2(e+f x) (g \tan (e+f x))^p\right ) \, dx}{a^4}\\ &=\frac {\int \sec ^4(e+f x) (g \tan (e+f x))^p \, dx}{a^2}+\frac {\int \sec ^2(e+f x) \tan ^2(e+f x) (g \tan (e+f x))^p \, dx}{a^2}-\frac {2 \int \sec ^3(e+f x) \tan (e+f x) (g \tan (e+f x))^p \, dx}{a^2}\\ &=\frac {\operatorname {Subst}\left (\int (g x)^p \left (1+x^2\right ) \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac {\int \sec ^2(e+f x) (g \tan (e+f x))^{2+p} \, dx}{a^2 g^2}-\frac {2 \int \sec ^3(e+f x) (g \tan (e+f x))^{1+p} \, dx}{a^2 g}\\ &=-\frac {2 \cos ^2(e+f x)^{\frac {5+p}{2}} \, _2F_1\left (\frac {2+p}{2},\frac {5+p}{2};\frac {4+p}{2};\sin ^2(e+f x)\right ) \sec ^3(e+f x) (g \tan (e+f x))^{2+p}}{a^2 f g^2 (2+p)}+\frac {\operatorname {Subst}\left (\int \left ((g x)^p+\frac {(g x)^{2+p}}{g^2}\right ) \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac {\operatorname {Subst}\left (\int (g x)^{2+p} \, dx,x,\tan (e+f x)\right )}{a^2 f g^2}\\ &=\frac {(g \tan (e+f x))^{1+p}}{a^2 f g (1+p)}-\frac {2 \cos ^2(e+f x)^{\frac {5+p}{2}} \, _2F_1\left (\frac {2+p}{2},\frac {5+p}{2};\frac {4+p}{2};\sin ^2(e+f x)\right ) \sec ^3(e+f x) (g \tan (e+f x))^{2+p}}{a^2 f g^2 (2+p)}+\frac {2 (g \tan (e+f x))^{3+p}}{a^2 f g^3 (3+p)}\\ \end {align*}

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Mathematica [B]  time = 13.96, size = 667, normalized size = 4.83 \[ \frac {2^{p+1} \tan \left (\frac {1}{2} (e+f x)\right ) \left (1-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^p \left (-\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\tan ^2\left (\frac {1}{2} (e+f x)\right )-1}\right )^p \tan ^{-p}(e+f x) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 (g \tan (e+f x))^p \left (\frac {\tan ^2\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 )}{p+3}-\frac {6 \tan ^2\left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (\frac {p+3}{2},p+3;\frac {p+5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+3}+\frac {12 \tan ^2\left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (\frac {p+3}{2},p+4;\frac {p+5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+3}-\frac {2 \tan \left (\frac {1}{2} (e+f x)\right ) \, _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}+\frac {6 \tan \left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (\frac {p+2}{2},p+3;\frac {p+4}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+2}-\frac {8 \tan \left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (\frac {p+2}{2},p+4;\frac {p+4}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+2}+\frac {\, _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}-\frac {2 \, _2F_1\left (\frac {p+1}{2},p+3;\frac {p+3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+1}+\frac {2 \, _2F_1\left (\frac {p+1}{2},p+4;\frac {p+3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+1}+\frac {2 \tan ^4\left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (p+4,\frac {p+5}{2};\frac {p+7}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+5}+\frac {2 \tan ^3\left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (p+3,\frac {p+4}{2};\frac {p+6}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+4}-\frac {8 \tan ^3\left (\frac {1}{2} (e+f x)\right ) \, _2F_1\left (\frac {p+4}{2},p+4;\frac {p+6}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{p+4}\right )}{f (a \sin (e+f x)+a)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*Tan[e + f*x])^p/(a + a*Sin[e + f*x])^2,x]

[Out]

(2^(1 + p)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Tan[(e + f*x)/2]*(1 - Tan[(e + f*x)/2]^2)^p*(-(Tan[(e + f*x
)/2]/(-1 + Tan[(e + f*x)/2]^2)))^p*(Hypergeometric2F1[(1 + p)/2, 2 + p, (3 + p)/2, Tan[(e + f*x)/2]^2]/(1 + p)
- (2*Hypergeometric2F1[(1 + p)/2, 3 + p, (3 + p)/2, Tan[(e + f*x)/2]^2])/(1 + p) + (2*Hypergeometric2F1[(1 +
p)/2, 4 + p, (3 + p)/2, Tan[(e + f*x)/2]^2])/(1 + p) - (2*Hypergeometric2F1[(2 + p)/2, 2 + p, (4 + p)/2, Tan[(
e + f*x)/2]^2]*Tan[(e + f*x)/2])/(2 + p) + (6*Hypergeometric2F1[(2 + p)/2, 3 + p, (4 + p)/2, Tan[(e + f*x)/2]^
2]*Tan[(e + f*x)/2])/(2 + p) - (8*Hypergeometric2F1[(2 + p)/2, 4 + p, (4 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e +
f*x)/2])/(2 + p) + (Hypergeometric2F1[2 + p, (3 + p)/2, (5 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)/(3
+ p) - (6*Hypergeometric2F1[(3 + p)/2, 3 + p, (5 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)/(3 + p) + (12
*Hypergeometric2F1[(3 + p)/2, 4 + p, (5 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^2)/(3 + p) + (2*Hypergeom
etric2F1[3 + p, (4 + p)/2, (6 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^3)/(4 + p) - (8*Hypergeometric2F1[(
4 + p)/2, 4 + p, (6 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^3)/(4 + p) + (2*Hypergeometric2F1[4 + p, (5 +
p)/2, (7 + p)/2, Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]^4)/(5 + p))*(g*Tan[e + f*x])^p)/(f*(a + a*Sin[e + f*x])
^2*Tan[e + f*x]^p)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\left (g \tan \left (f x + e\right )\right )^{p}}{a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(-(g*tan(f*x + e))^p/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((g*tan(f*x + e))^p/(a*sin(f*x + e) + a)^2, x)

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maple [F]  time = 1.02, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \tan \left (f x +e \right )\right )^{p}}{\left (a +a \sin \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (g \tan \left (f x + e\right )\right )^{p}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))^p/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((g*tan(f*x + e))^p/(a*sin(f*x + e) + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (g\,\mathrm {tan}\left (e+f\,x\right )\right )}^p}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (g \tan {\left (e + f x \right )}\right )^{p}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*tan(f*x+e))**p/(a+a*sin(f*x+e))**2,x)

[Out]

Integral((g*tan(e + f*x))**p/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1), x)/a**2

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