3.498 \(\int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx\)

Optimal. Leaf size=73 \[ -\frac{b \sin (e+f x) (b \sec (e+f x))^{n-1} \, _2F_1\left (-\frac{3}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f (1-n) \sqrt{\sin ^2(e+f x)}} \]

[Out]

-((b*Hypergeometric2F1[-3/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Sec[e + f*x])^(-1 + n)*Sin[e + f*x])/(f*
(1 - n)*Sqrt[Sin[e + f*x]^2]))

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Rubi [A]  time = 0.0818204, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2632, 2576} \[ -\frac{b \sin (e+f x) (b \sec (e+f x))^{n-1} \, _2F_1\left (-\frac{3}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f (1-n) \sqrt{\sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^n*Sin[e + f*x]^4,x]

[Out]

-((b*Hypergeometric2F1[-3/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Sec[e + f*x])^(-1 + n)*Sin[e + f*x])/(f*
(1 - n)*Sqrt[Sin[e + f*x]^2]))

Rule 2632

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^2*(a*Sec[e
+ f*x])^(m - 1)*(b*Csc[e + f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/b^2, Int[1/((a*Co
s[e + f*x])^m*(b*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x]

Rule 2576

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^(2*IntPar
t[(n - 1)/2] + 1)*(b*Sin[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Cos[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/
2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2])/(a*f*(m + 1)*(Sin[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a,
b, e, f, m, n}, x] && SimplerQ[n, m]

Rubi steps

\begin{align*} \int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx &=\left (b^2 (b \cos (e+f x))^{-1+n} (b \sec (e+f x))^{-1+n}\right ) \int (b \cos (e+f x))^{-n} \sin ^4(e+f x) \, dx\\ &=-\frac{b \, _2F_1\left (-\frac{3}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sin (e+f x)}{f (1-n) \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 24.7412, size = 6192, normalized size = 84.82 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x]^4,x]

[Out]

Result too large to show

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Maple [F]  time = 0.701, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( fx+e \right ) \right ) ^{n} \left ( \sin \left ( fx+e \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^n*sin(f*x+e)^4,x)

[Out]

int((b*sec(f*x+e))^n*sin(f*x+e)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^n*sin(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \left (b \sec \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(b*sec(f*x + e))^n, 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((b*sec(f*x+e))**n*sin(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 (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^n*sin(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n*sin(f*x + e)^4, x)