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*hypergeom([-3/2, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*sec(f*x+e))^(-1+n)*sin(f*x+e)/f/(1-n)/(sin(f*x+e)^
2)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 24.18, 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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \left (b \sec \left (f x + e\right )\right )^{n}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4}\,{d x} \]

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)

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maple [F]  time = 1.06, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x +e \right )\right )^{n} \left (\sin ^{4}\left (f x +e \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{4}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (e + f x \right )}\right )^{n} \sin ^{4}{\left (e + f x \right )}\, dx \]

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]

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

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