3.205 \(\int (a+b \sin (e+f x)) (g \tan (e+f x))^p \, dx\)

Optimal. Leaf size=129 \[ \frac{a (g \tan (e+f x))^{p+1} \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};-\tan ^2(e+f x)\right )}{f g (p+1)}+\frac{b \sin (e+f x) \cos ^2(e+f x)^{\frac{p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac{p+1}{2},\frac{p+2}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{f g (p+2)} \]

[Out]

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

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Rubi [A]  time = 0.147865, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2722, 3476, 364, 2602, 2577} \[ \frac{a (g \tan (e+f x))^{p+1} \, _2F_1\left (1,\frac{p+1}{2};\frac{p+3}{2};-\tan ^2(e+f x)\right )}{f g (p+1)}+\frac{b \sin (e+f x) \cos ^2(e+f x)^{\frac{p+1}{2}} (g \tan (e+f x))^{p+1} \, _2F_1\left (\frac{p+1}{2},\frac{p+2}{2};\frac{p+4}{2};\sin ^2(e+f x)\right )}{f g (p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2722

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^
n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

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

Mathematica [C]  time = 8.26249, size = 849, normalized size = 6.58 \[ \frac{2 (a+b \sin (e+f x)) \tan \left (\frac{1}{2} (e+f x)\right ) \left (a (p+2) F_1\left (\frac{p+1}{2};p,1;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 b (p+1) F_1\left (\frac{p}{2}+1;p,2;\frac{p}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) \tan \left (\frac{1}{2} (e+f x)\right )\right ) (g \tan (e+f x))^p}{f \left (-16 p \cos \left (\frac{1}{2} (e+f x)\right ) \csc ^3(e+f x) \sec (e+f x) \left (a (p+2) F_1\left (\frac{p+1}{2};p,1;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 b (p+1) F_1\left (\frac{p}{2}+1;p,2;\frac{p}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) \tan \left (\frac{1}{2} (e+f x)\right )\right ) \sin ^5\left (\frac{1}{2} (e+f x)\right )+\sec ^2\left (\frac{1}{2} (e+f x)\right ) \left (a (p+2) F_1\left (\frac{p+1}{2};p,1;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 b (p+1) F_1\left (\frac{p}{2}+1;p,2;\frac{p}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) \tan \left (\frac{1}{2} (e+f x)\right )\right )+2 p \csc (e+f x) \sec (e+f x) \tan \left (\frac{1}{2} (e+f x)\right ) \left (a (p+2) F_1\left (\frac{p+1}{2};p,1;\frac{p+3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )+2 b (p+1) F_1\left (\frac{p}{2}+1;p,2;\frac{p}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right ) \tan \left (\frac{1}{2} (e+f x)\right )\right )+2 (p+1) \sec ^2\left (\frac{1}{2} (e+f x)\right ) \tan \left (\frac{1}{2} (e+f x)\right ) \left (\frac{2 b (p+2) \left (p F_1\left (\frac{p}{2}+2;p+1,2;\frac{p}{2}+3;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 F_1\left (\frac{p}{2}+2;p,3;\frac{p}{2}+3;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac{1}{2} (e+f x)\right )}{p+4}+\frac{a (p+2) \left (p F_1\left (\frac{p+3}{2};p+1,1;\frac{p+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-F_1\left (\frac{p+3}{2};p,2;\frac{p+5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right ) \tan \left (\frac{1}{2} (e+f x)\right )}{p+3}+b F_1\left (\frac{p}{2}+1;p,2;\frac{p}{2}+2;\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(a + b*Sin[e + f*x])*Tan[(e + f*x)/2]*(a*(2 + p)*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan[(e + f*x)/2]^2, -
Tan[(e + f*x)/2]^2] + 2*b*(1 + p)*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Ta
n[(e + f*x)/2])*(g*Tan[e + f*x])^p)/(f*(Sec[(e + f*x)/2]^2*(a*(2 + p)*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*b*(1 + p)*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[
(e + f*x)/2]^2]*Tan[(e + f*x)/2]) - 16*p*Cos[(e + f*x)/2]*Csc[e + f*x]^3*Sec[e + f*x]*Sin[(e + f*x)/2]^5*(a*(2
+ p)*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*b*(1 + p)*AppellF1[1 +
p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]) + 2*p*Csc[e + f*x]*Sec[e + f*x
]*Tan[(e + f*x)/2]*(a*(2 + p)*AppellF1[(1 + p)/2, p, 1, (3 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] +
2*b*(1 + p)*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Tan[(e + f*x)/2]) + 2*(1
+ p)*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]*(b*AppellF1[1 + p/2, p, 2, 2 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2] + (a*(2 + p)*(-AppellF1[(3 + p)/2, p, 2, (5 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + p*Appe
llF1[(3 + p)/2, 1 + p, 1, (5 + p)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2])/(3 + p) + (2*
b*(2 + p)*(-2*AppellF1[2 + p/2, p, 3, 3 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + p*AppellF1[2 + p/2,
1 + p, 2, 3 + p/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/(4 + p))))

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Maple [F]  time = 0.904, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sin \left ( fx+e \right ) \right ) \left ( g\tan \left ( fx+e \right ) \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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