3.5.98 \(\int (b \sec (e+f x))^n \sin ^4(e+f x) \, dx\) [498]

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

[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)

________________________________________________________________________________________

Rubi [A]
time = 0.06, 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 = {2712, 2656} \begin {gather*} -\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)}} \end {gather*}

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 2656

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)/(a*f*(m + 1)*(Sin[e + f*
x]^2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2], x] /; FreeQ[{a
, b, e, f, m, n}, x] && SimplerQ[n, m]

Rule 2712

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^2/b^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), 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*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 22.01, size = 6192, normalized size = 84.82 \begin {gather*} \text {Result too large to show} \end {gather*}

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

________________________________________________________________________________________

Maple [F]
time = 0.24, 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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________