3.2.31 \(\int (a+a \sin (e+f x))^m \tan (e+f x) \, dx\) [131]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 = {2786, 80, 70} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a + a*Sin[e + f*x])^m/(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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 80

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 2786

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]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m \tan (e+f x) \, dx &=\frac {\text {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 {\text {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*}

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Mathematica [A]
time = 0.05, size = 63, normalized size = 0.88 \begin {gather*} \frac {(a (1+\sin (e+f x)))^m \left (-2 (1+m)+m \, _2F_1\left (1,1+m;2+m;\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )}{4 f m (1+m)} \end {gather*}

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 = 0.12, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \tan \left (f x +e \right )\, dx\]

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 = 0.00, 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+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 = 0.00, 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+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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \tan {\left (e + f x \right )}\, dx \end {gather*}

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 = 0.00, 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+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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {tan}\left (e+f\,x\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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