3.135 \(\int (a+a \sin (e+f x))^m \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=311 \[ -\frac{a^2 \sin ^2(e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac{a^2 \sin (e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f (1-m) (a-a \sin (e+f x))}+\frac{2^{m-\frac{3}{2}} \left (m^4+6 m^3-7 m^2-12 m+9\right ) (1-\sin (e+f x)) \sec (e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{5}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f (1-m) m}-\frac{\sec (e+f x) (a \sin (e+f x)+a)^{m-1} \left (a \left (-m^3-7 m^2-m+6\right )-a \left (-m^3-8 m^2-6 m+9\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))} \]

[Out]

(2^(-3/2 + m)*(9 - 12*m - 7*m^2 + 6*m^3 + m^4)*Hypergeometric2F1[1/2, 5/2 - m, 3/2, (1 - Sin[e + f*x])/2]*Sec[
e + f*x]*(1 - Sin[e + f*x])*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^m)/(3*f*(1 - m)*m) - (Sec[e + f*
x]*(a + a*Sin[e + f*x])^(-1 + m)*(a*(6 - m - 7*m^2 - m^3) - a*(9 - 6*m - 8*m^2 - m^3)*Sin[e + f*x]))/(3*f*(1 -
m)*m*(1 - Sin[e + f*x])) + (a^2*Sin[e + f*x]*(a + a*Sin[e + f*x])^(-1 + m)*Tan[e + f*x])/(f*(1 - m)*(a - a*Si
n[e + f*x])) - (a^2*Sin[e + f*x]^2*(a + a*Sin[e + f*x])^(-1 + m)*Tan[e + f*x])/(f*m*(a - a*Sin[e + f*x]))

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Rubi [A]  time = 0.355058, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2719, 100, 153, 145, 70, 69} \[ -\frac{a^2 \sin ^2(e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f m (a-a \sin (e+f x))}+\frac{a^2 \sin (e+f x) \tan (e+f x) (a \sin (e+f x)+a)^{m-1}}{f (1-m) (a-a \sin (e+f x))}+\frac{2^{m-\frac{3}{2}} \left (m^4+6 m^3-7 m^2-12 m+9\right ) (1-\sin (e+f x)) \sec (e+f x) (\sin (e+f x)+1)^{\frac{1}{2}-m} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{5}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{3 f (1-m) m}-\frac{\sec (e+f x) (a \sin (e+f x)+a)^{m-1} \left (a \left (-m^3-7 m^2-m+6\right )-a \left (-m^3-8 m^2-6 m+9\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*Tan[e + f*x]^4,x]

[Out]

(2^(-3/2 + m)*(9 - 12*m - 7*m^2 + 6*m^3 + m^4)*Hypergeometric2F1[1/2, 5/2 - m, 3/2, (1 - Sin[e + f*x])/2]*Sec[
e + f*x]*(1 - Sin[e + f*x])*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^m)/(3*f*(1 - m)*m) - (Sec[e + f*
x]*(a + a*Sin[e + f*x])^(-1 + m)*(a*(6 - m - 7*m^2 - m^3) - a*(9 - 6*m - 8*m^2 - m^3)*Sin[e + f*x]))/(3*f*(1 -
m)*m*(1 - Sin[e + f*x])) + (a^2*Sin[e + f*x]*(a + a*Sin[e + f*x])^(-1 + m)*Tan[e + f*x])/(f*(1 - m)*(a - a*Si
n[e + f*x])) - (a^2*Sin[e + f*x]^2*(a + a*Sin[e + f*x])^(-1 + m)*Tan[e + f*x])/(f*m*(a - a*Sin[e + f*x]))

Rule 2719

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[(Sqrt[a + b*Si
n[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])/(b*f*Cos[e + f*x]), Subst[Int[(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[m] && Inte
gerQ[p/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 153

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

Rule 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

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 ^4(e+f x) \, dx &=\frac{\left (\sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^4 (a+x)^{-\frac{5}{2}+m}}{(a-x)^{5/2}} \, dx,x,a \sin (e+f x)\right )}{a f}\\ &=-\frac{a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac{\left (\sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2 (a+x)^{-\frac{5}{2}+m} \left (-3 a^2-a m x\right )}{(a-x)^{5/2}} \, dx,x,a \sin (e+f x)\right )}{a f m}\\ &=\frac{a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac{a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac{\left (\sec (e+f x) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x (a+x)^{-\frac{5}{2}+m} \left (2 a^3 m-a^2 \left (3-3 m-m^2\right ) x\right )}{(a-x)^{5/2}} \, dx,x,a \sin (e+f x)\right )}{a f (1-m) m}\\ &=-\frac{\sec (e+f x) (a+a \sin (e+f x))^{-1+m} \left (a \left (6-m-7 m^2-m^3\right )-a \left (9-6 m-8 m^2-m^3\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))}+\frac{a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac{a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac{\left (a \left (9-12 m-7 m^2+6 m^3+m^4\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+x)^{-\frac{5}{2}+m}}{\sqrt{a-x}} \, dx,x,a \sin (e+f x)\right )}{3 f (1-m) m}\\ &=-\frac{\sec (e+f x) (a+a \sin (e+f x))^{-1+m} \left (a \left (6-m-7 m^2-m^3\right )-a \left (9-6 m-8 m^2-m^3\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))}+\frac{a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac{a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}-\frac{\left (2^{-\frac{5}{2}+m} \left (9-12 m-7 m^2+6 m^3+m^4\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 a}\right )^{-\frac{5}{2}+m}}{\sqrt{a-x}} \, dx,x,a \sin (e+f x)\right )}{3 a f (1-m) m}\\ &=\frac{2^{-\frac{3}{2}+m} \left (9-12 m-7 m^2+6 m^3+m^4\right ) \, _2F_1\left (\frac{1}{2},\frac{5}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1-\sin (e+f x)) (1+\sin (e+f x))^{\frac{1}{2}-m} (a+a \sin (e+f x))^m}{3 f (1-m) m}-\frac{\sec (e+f x) (a+a \sin (e+f x))^{-1+m} \left (a \left (6-m-7 m^2-m^3\right )-a \left (9-6 m-8 m^2-m^3\right ) \sin (e+f x)\right )}{3 f (1-m) m (1-\sin (e+f x))}+\frac{a^2 \sin (e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f (1-m) (a-a \sin (e+f x))}-\frac{a^2 \sin ^2(e+f x) (a+a \sin (e+f x))^{-1+m} \tan (e+f x)}{f m (a-a \sin (e+f x))}\\ \end{align*}

Mathematica [F]  time = 1.06385, size = 0, normalized size = 0. \[ \int (a+a \sin (e+f x))^m \tan ^4(e+f x) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + a*Sin[e + f*x])^m*Tan[e + f*x]^4,x]

[Out]

Integrate[(a + a*Sin[e + f*x])^m*Tan[e + f*x]^4, x]

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Maple [F]  time = 0.152, 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 ) ^{4}\, 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)^4,x)

[Out]

int((a+a*sin(f*x+e))^m*tan(f*x+e)^4,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 )^{4}\,{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)^4,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^4, 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 )^{4}, 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)^4,x, algorithm="fricas")

[Out]

integral((a*sin(f*x + e) + a)^m*tan(f*x + e)^4, 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)**4,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 )^{4}\,{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)^4,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^m*tan(f*x + e)^4, x)