3.131 \(\int (a+a \sin (e+f x))^m \tan (e+f x) \, dx\)

Optimal. Leaf size=72 \[ \frac{(a \sin (e+f x)+a)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{1}{2} (\sin (e+f x)+1)\right )}{4 a f (m+1)}-\frac{(a \sin (e+f x)+a)^m}{2 f m} \]

[Out]

-(a + a*Sin[e + f*x])^m/(2*f*m) + (Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x
])^(1 + m))/(4*a*f*(1 + m))

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Rubi [A]  time = 0.0504618, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2707, 79, 68} \[ \frac{(a \sin (e+f x)+a)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{1}{2} (\sin (e+f x)+1)\right )}{4 a f (m+1)}-\frac{(a \sin (e+f x)+a)^m}{2 f m} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*Tan[e + f*x],x]

[Out]

-(a + a*Sin[e + f*x])^m/(2*f*m) + (Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x
])^(1 + m))/(4*a*f*(1 + m))

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.0706524, size = 63, normalized size = 0.88 \[ \frac{(a (\sin (e+f x)+1))^m \left (m (\sin (e+f x)+1) \, _2F_1\left (1,m+1;m+2;\frac{1}{2} (\sin (e+f x)+1)\right )-2 (m+1)\right )}{4 f m (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^m*Tan[e + f*x],x]

[Out]

((a*(1 + Sin[e + f*x]))^m*(-2*(1 + m) + m*Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[e + f*x])/2]*(1 + Sin[e
+ f*x])))/(4*f*m*(1 + m))

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Maple [F]  time = 0.76, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\tan \left ( fx+e \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*tan(f*x+e),x)

[Out]

int((a+a*sin(f*x+e))^m*tan(f*x+e),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sin(f*x + e) + a)^m*tan(f*x + e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**m*tan(f*x+e),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*tan(e + f*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e), x)