3.130 \(\int (a+a \sin (e+f x))^m \tan ^3(e+f x) \, dx\)

Optimal. Leaf size=163 \[ -\frac{a^2 \sin ^2(e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac{a (m+4) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac{1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)}+\frac{\left (2 a m \sin (e+f x)+a \left (-m^2-3 m+2\right )\right ) (a \sin (e+f x)+a)^{m-1}}{2 f (1-m) m (1-\sin (e+f x))} \]

[Out]

(a*(4 + m)*Hypergeometric2F1[1, -1 + m, m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^(-1 + m))/(4*f*(1 - m))
- (a^2*Sin[e + f*x]^2*(a + a*Sin[e + f*x])^(-1 + m))/(f*m*(a - a*Sin[e + f*x])) + ((a + a*Sin[e + f*x])^(-1 +
m)*(a*(2 - 3*m - m^2) + 2*a*m*Sin[e + f*x]))/(2*f*(1 - m)*m*(1 - Sin[e + f*x]))

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Rubi [A]  time = 0.148753, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2707, 100, 146, 68} \[ -\frac{a^2 \sin ^2(e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac{a (m+4) (a \sin (e+f x)+a)^{m-1} \, _2F_1\left (1,m-1;m;\frac{1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)}+\frac{\left (2 a m \sin (e+f x)+a \left (-m^2-3 m+2\right )\right ) (a \sin (e+f x)+a)^{m-1}}{2 f (1-m) m (1-\sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*Tan[e + f*x]^3,x]

[Out]

(a*(4 + m)*Hypergeometric2F1[1, -1 + m, m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^(-1 + m))/(4*f*(1 - m))
- (a^2*Sin[e + f*x]^2*(a + a*Sin[e + f*x])^(-1 + m))/(f*m*(a - a*Sin[e + f*x])) + ((a + a*Sin[e + f*x])^(-1 +
m)*(a*(2 - 3*m - m^2) + 2*a*m*Sin[e + f*x]))/(2*f*(1 - m)*m*(1 - Sin[e + f*x]))

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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

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

Mathematica [A]  time = 0.267321, size = 105, normalized size = 0.64 \[ \frac{a (a (\sin (e+f x)+1))^{m-1} \left (-m (m+4) (\sin (e+f x)-1) \, _2F_1\left (1,m-1;m;\frac{1}{2} (\sin (e+f x)+1)\right )+4 (m-1) \sin ^2(e+f x)+4 m \sin (e+f x)-2 \left (m^2+3 m-2\right )\right )}{4 f (m-1) m (\sin (e+f x)-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^m*Tan[e + f*x]^3,x]

[Out]

(a*(a*(1 + Sin[e + f*x]))^(-1 + m)*(-2*(-2 + 3*m + m^2) - m*(4 + m)*Hypergeometric2F1[1, -1 + m, m, (1 + Sin[e
+ f*x])/2]*(-1 + Sin[e + f*x]) + 4*m*Sin[e + f*x] + 4*(-1 + m)*Sin[e + f*x]^2))/(4*f*(-1 + m)*m*(-1 + Sin[e +
f*x]))

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Maple [F]  time = 0.144, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( \tan \left ( fx+e \right ) \right ) ^{3}\, 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)^3,x)

[Out]

int((a+a*sin(f*x+e))^m*tan(f*x+e)^3,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 )^{3}\,{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)^3,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^3, 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 )^{3}, 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)^3,x, algorithm="fricas")

[Out]

integral((a*sin(f*x + e) + a)^m*tan(f*x + e)^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**3,x)

[Out]

Timed out

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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 )^{3}\,{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)^3,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^3, x)