3.386 \(\int (b \sec (e+f x))^{3/2} \sin ^5(e+f x) \, dx\)

Optimal. Leaf size=63 \[ -\frac {2 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {4 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[Out]

-2/7*b^5/f/(b*sec(f*x+e))^(7/2)+4/3*b^3/f/(b*sec(f*x+e))^(3/2)+2*b*(b*sec(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.06, 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 = {2622, 270} \[ -\frac {2 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {4 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^5,x]

[Out]

(-2*b^5)/(7*f*(b*Sec[e + f*x])^(7/2)) + (4*b^3)/(3*f*(b*Sec[e + f*x])^(3/2)) + (2*b*Sqrt[b*Sec[e + f*x]])/f

Rule 270

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 2622

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))^{3/2} \sin ^5(e+f x) \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {\left (-1+\frac {x^2}{b^2}\right )^2}{x^{9/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {b^5 \operatorname {Subst}\left (\int \left (\frac {1}{x^{9/2}}-\frac {2}{b^2 x^{5/2}}+\frac {1}{b^4 \sqrt {x}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=-\frac {2 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {4 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 42, normalized size = 0.67 \[ \frac {b (44 \cos (2 (e+f x))-3 \cos (4 (e+f x))+215) \sqrt {b \sec (e+f x)}}{84 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^5,x]

[Out]

(b*(215 + 44*Cos[2*(e + f*x)] - 3*Cos[4*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(84*f)

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fricas [A]  time = 0.66, size = 43, normalized size = 0.68 \[ -\frac {2 \, {\left (3 \, b \cos \left (f x + e\right )^{4} - 14 \, b \cos \left (f x + e\right )^{2} - 21 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{21 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x, algorithm="fricas")

[Out]

-2/21*(3*b*cos(f*x + e)^4 - 14*b*cos(f*x + e)^2 - 21*b)*sqrt(b/cos(f*x + e))/f

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sin \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(3/2)*sin(f*x + e)^5, x)

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maple [B]  time = 0.23, size = 959, normalized size = 15.22 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x)

[Out]

-1/42/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))^2*(-21*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)+21*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)-63*cos(f*
x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)+63*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f
*x+e)+1)^2)^(3/2)*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)+12*cos(f*x+e)^5-63*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3
/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)+63*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*ln(-(2*cos(f*x+e)^2*(-co
s(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)-21*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)/(c
os(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+21*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)/(cos(f*x+e)+1)^2)^(3/2)-56*cos(f*x+e)^3-84*cos(f*x+e))*(b/cos(f*x+e))^(3/2)/sin(f*x+e)^4

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maxima [A]  time = 0.42, size = 55, normalized size = 0.87 \[ -\frac {2 \, b {\left (\frac {3 \, b^{4}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}} - \frac {14 \, b^{2}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} - 21 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{21 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^5,x, algorithm="maxima")

[Out]

-2/21*b*(3*b^4/(b/cos(f*x + e))^(7/2) - 14*b^2/(b/cos(f*x + e))^(3/2) - 21*sqrt(b/cos(f*x + e)))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\sin \left (e+f\,x\right )}^5\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5*(b/cos(e + f*x))^(3/2),x)

[Out]

int(sin(e + f*x)^5*(b/cos(e + f*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**(3/2)*sin(f*x+e)**5,x)

[Out]

Timed out

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