3.373 \(\int \sqrt {b \sec (e+f x)} \sin ^3(e+f x) \, dx\)

Optimal. Leaf size=41 \[ \frac {2 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.04, 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} \[ \frac {2 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3,x]

[Out]

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

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

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Mathematica [A]  time = 0.16, size = 36, normalized size = 0.88 \[ \frac {(\cos (3 (e+f x))-17 \cos (e+f x)) \sqrt {b \sec (e+f x)}}{10 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^3,x]

[Out]

((-17*Cos[e + f*x] + Cos[3*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(10*f)

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fricas [A]  time = 0.69, size = 34, normalized size = 0.83 \[ \frac {2 \, {\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{5 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)-16/5*(5*b^2*sqrt(-b)*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))-5*b*sqrt(-b)*
(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^3-5*b^2*(-sqrt(-b)*tan(1/2*(f*x+exp(1))
)^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^2+b^3)*sign(cos(f*x+exp(1)))/(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-
b*tan(1/2*(f*x+exp(1)))^4+b)-sqrt(-b))^5/f

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maple [B]  time = 0.21, size = 497, normalized size = 12.12 \[ \frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (4 \left (\cos ^{3}\left (f x +e \right )\right )-5 \cos \left (f x +e \right ) \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 )+5 \cos \left (f x +e \right ) \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 )-5 \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 )+5 \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 )-20 \cos \left (f x +e \right )\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {\frac {b}{\cos \left (f x +e \right )}}}{10 f \sin \left (f x +e \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x)

[Out]

1/10/f*(-1+cos(f*x+e))^2*(4*cos(f*x+e)^3-5*cos(f*x+e)*(-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)+5*cos(f*x+e)*(-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)-5*(-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)+5*(-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)-20*cos(f*x+e))*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(1/2)/sin(f*x+e)^4

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maxima [A]  time = 0.39, size = 35, normalized size = 0.85 \[ \frac {2 \, {\left (b^{2} - \frac {5 \, b^{2}}{\cos \left (f x + e\right )^{2}}\right )} b}{5 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^3*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/5*(b^2 - 5*b^2/cos(f*x + e)^2)*b/(f*(b/cos(f*x + e))^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^3*(b/cos(e + f*x))^(1/2),x)

[Out]

int(sin(e + f*x)^3*(b/cos(e + f*x))^(1/2), 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(sin(f*x+e)**3*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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