3.136 \(\int (a+a \sin (e+f x))^m \tan ^2(e+f x) \, dx\)

Optimal. Leaf size=157 \[ \frac{2^{m-\frac{1}{2}} \left (-m^2-m+1\right ) \sec (e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (-\frac{1}{2},\frac{3}{2}-m;\frac{1}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (1-m) m}+\frac{\sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m) m}-\frac{\sec (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f m} \]

[Out]

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

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Rubi [A]  time = 0.248016, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2713, 2860, 2689, 70, 69} \[ \frac{2^{m-\frac{1}{2}} \left (-m^2-m+1\right ) \sec (e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (-\frac{1}{2},\frac{3}{2}-m;\frac{1}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{f (1-m) m}+\frac{\sec (e+f x) (a \sin (e+f x)+a)^m}{f (1-m) m}-\frac{\sec (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2713

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

Rule 2860

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rule 2689

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

Rule 70

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

Rule 69

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

Rubi steps

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

Mathematica [C]  time = 34.458, size = 11184, normalized size = 71.24 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.12, 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 ) ^{2}\, 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)^2,x)

[Out]

int((a+a*sin(f*x+e))^m*tan(f*x+e)^2,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 )^{2}\,{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)^2,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^2, 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 )^{2}, 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)^2,x, algorithm="fricas")

[Out]

integral((a*sin(f*x + e) + a)^m*tan(f*x + e)^2, 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)**2,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 )^{2}\,{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)^2,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^2, x)