3.4.98 \(\int (b \sec (e+f x))^{5/2} \sin ^5(e+f x) \, dx\) [398]

Optimal. Leaf size=63 \[ -\frac {2 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac {4 b^3}{f \sqrt {b \sec (e+f x)}}+\frac {2 b (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

-2/5*b^5/f/(b*sec(f*x+e))^(5/2)+2/3*b*(b*sec(f*x+e))^(3/2)/f+4*b^3/f/(b*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2702, 276} \begin {gather*} -\frac {2 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac {4 b^3}{f \sqrt {b \sec (e+f x)}}+\frac {2 b (b \sec (e+f x))^{3/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^5,x]

[Out]

(-2*b^5)/(5*f*(b*Sec[e + f*x])^(5/2)) + (4*b^3)/(f*Sqrt[b*Sec[e + f*x]]) + (2*b*(b*Sec[e + f*x])^(3/2))/(3*f)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int (b \sec (e+f x))^{5/2} \sin ^5(e+f x) \, dx &=\frac {b^5 \text {Subst}\left (\int \frac {\left (-1+\frac {x^2}{b^2}\right )^2}{x^{7/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {b^5 \text {Subst}\left (\int \left (\frac {1}{x^{7/2}}-\frac {2}{b^2 x^{3/2}}+\frac {\sqrt {x}}{b^4}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac {2 b^5}{5 f (b \sec (e+f x))^{5/2}}+\frac {4 b^3}{f \sqrt {b \sec (e+f x)}}+\frac {2 b (b \sec (e+f x))^{3/2}}{3 f}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 42, normalized size = 0.67 \begin {gather*} \frac {b (151+108 \cos (2 (e+f x))-3 \cos (4 (e+f x))) (b \sec (e+f x))^{3/2}}{60 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(5/2)*Sin[e + f*x]^5,x]

[Out]

(b*(151 + 108*Cos[2*(e + f*x)] - 3*Cos[4*(e + f*x)])*(b*Sec[e + f*x])^(3/2))/(60*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(521\) vs. \(2(53)=106\).
time = 0.26, size = 522, normalized size = 8.29

method	result	size
default	\(-\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (6 \left (\cos ^{4}\left (f x +e \right )\right )+15 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-15 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+15 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right ) \cos \left (f x +e \right )-15 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right ) \cos \left (f x +e \right )-60 \left (\cos ^{2}\left (f x +e \right )\right )-10\right ) \cos \left (f x +e \right ) \left (\cos \left (f x +e \right )+1\right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{15 f \sin \left (f x +e \right )^{4}}\)	\(522\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(5/2)*sin(f*x+e)^5,x,method=_RETURNVERBOSE)

[Out]

-1/15/f*(-1+cos(f*x+e))^2*(6*cos(f*x+e)^4+15*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(
f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^
2)*cos(f*x+e)^2-15*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/
2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*cos(f*x+e)^2+15*(-cos(f*x
+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+
e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*cos(f*x+e)-15*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*
ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(1/2)-1)/sin(f*x+e)^2)*cos(f*x+e)-60*cos(f*x+e)^2-10)*cos(f*x+e)*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(5/2)/s
in(f*x+e)^4

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Maxima [A]
time = 0.30, size = 55, normalized size = 0.87 \begin {gather*} \frac {2 \, {\left (5 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}} - \frac {3 \, {\left (b^{4} - \frac {10 \, b^{4}}{\cos \left (f x + e\right )^{2}}\right )}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {5}{2}}}\right )} b}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^5,x, algorithm="maxima")

[Out]

2/15*(5*(b/cos(f*x + e))^(3/2) - 3*(b^4 - 10*b^4/cos(f*x + e)^2)/(b/cos(f*x + e))^(5/2))*b/f

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Fricas [A]
time = 0.42, size = 61, normalized size = 0.97 \begin {gather*} -\frac {2 \, {\left (3 \, b^{2} \cos \left (f x + e\right )^{4} - 30 \, b^{2} \cos \left (f x + e\right )^{2} - 5 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^5,x, algorithm="fricas")

[Out]

-2/15*(3*b^2*cos(f*x + e)^4 - 30*b^2*cos(f*x + e)^2 - 5*b^2)*sqrt(b/cos(f*x + e))/(f*cos(f*x + e))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**(5/2)*sin(f*x+e)**5,x)

[Out]

Timed out

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Giac [A]
time = 3.55, size = 80, normalized size = 1.27 \begin {gather*} -\frac {2 \, {\left (3 \, \sqrt {b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - 30 \, \sqrt {b \cos \left (f x + e\right )} b^{2} - \frac {5 \, b^{3}}{\sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(5/2)*sin(f*x+e)^5,x, algorithm="giac")

[Out]

-2/15*(3*sqrt(b*cos(f*x + e))*b^2*cos(f*x + e)^2 - 30*sqrt(b*cos(f*x + e))*b^2 - 5*b^3/(sqrt(b*cos(f*x + e))*c
os(f*x + e)))*sgn(cos(f*x + e))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int {\sin \left (e+f\,x\right )}^5\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5*(b/cos(e + f*x))^(5/2),x)

[Out]

int(sin(e + f*x)^5*(b/cos(e + f*x))^(5/2), x)

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