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

Optimal. Leaf size=41 \[ \frac {2 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[Out]

2/3*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 = 0.05, antiderivative size = 41, 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, 14} \[ \frac {2 b^3}{3 f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 ^3(e+f x) \, dx &=\frac {b^3 \operatorname {Subst}\left (\int \frac {-1+\frac {x^2}{b^2}}{x^{5/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {b^3 \operatorname {Subst}\left (\int \left (-\frac {1}{x^{5/2}}+\frac {1}{b^2 \sqrt {x}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {2 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 = 0.07, size = 30, normalized size = 0.73 \[ \frac {b (\cos (2 (e+f x))+7) \sqrt {b \sec (e+f x)}}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(7 + Cos[2*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(3*f)

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fricas [A]  time = 0.57, size = 31, normalized size = 0.76 \[ \frac {2 \, {\left (b \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*cos(f*x + e)^2 + 3*b)*sqrt(b/cos(f*x + e))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.19, size = 949, normalized size = 23.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))^2*(3*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)-3*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)+9*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)-9*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)+9*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)-9*cos(f*x+e)*(-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)+3*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)/s
in(f*x+e)^2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)-3*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-c
os(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)+4*cos(f*x+e)^3+12*cos(f*x+e))*(b/cos(f*x+e))^(3/2)/sin(f*x+e)^4

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maxima [A]  time = 0.36, size = 37, normalized size = 0.90 \[ \frac {2 \, b {\left (\frac {b^{2}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} + 3 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*b*(b^2/(b/cos(f*x + e))^(3/2) + 3*sqrt(b/cos(f*x + e)))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sin(e + f*x)^3*(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)**3,x)

[Out]

Timed out

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