∫ sin3 (e+fx)___∘ a+b sec2(e+fx) dx

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]

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

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Rubi [A] time = 0.09, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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 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])

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*}

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Mathematica [A] time = 0.28, 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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fricas [A] time = 0.50, size = 57, normalized size = 0.77 \[ \frac {{\left (a \cos \left (f x + e\right )^{3} - {\left (3 \, a + 2 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, a^{2} f} \]

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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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/(a+b*sec(f*x+e)^2)^(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)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si