3.94 \(\int \frac{\sin ^3(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=74 \[ \frac{\cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{3 a f}-\frac{(3 a+2 b) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{3 a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.0910306, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4134, 453, 264} \[ \frac{\cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{3 a f}-\frac{(3 a+2 b) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.294836, size = 64, normalized size = 0.86 \[ \frac{\sec (e+f x) (a \cos (2 (e+f x))-5 a-4 b) (a \cos (2 (e+f x))+a+2 b)}{12 a^2 f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*a - 4*b + a*Cos[2*(e + f*x)])*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x])/(12*a^2*f*Sqrt[a + b*Sec[e + f
*x]^2])

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Maple [A]  time = 0.344, size = 69, normalized size = 0.9 \begin{align*}{\frac{ \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}-3\,a-2\,b \right ) }{3\,f{a}^{2}\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/f/a^2*(b+a*cos(f*x+e)^2)*(a*cos(f*x+e)^2-3*a-2*b)/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)

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Maxima [A]  time = 1.02358, size = 112, normalized size = 1.51 \begin{align*} -\frac{\frac{3 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a} - \frac{{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{2}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a - ((a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 3*sqrt(a +
b/cos(f*x + e)^2)*b*cos(f*x + e))/a^2)/f

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Fricas [A]  time = 0.550606, size = 139, normalized size = 1.88 \begin{align*} \frac{{\left (a \cos \left (f x + e\right )^{3} -{\left (3 \, a + 2 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, a^{2} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*cos(f*x + e)^3 - (3*a + 2*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**3/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{3}}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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