3.2.44 \(\int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [144]

Optimal. Leaf size=83 \[ \frac {(a-b) \tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 b^{3/2} f}-\frac {\cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 b f} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3265, 396, 223, 209} \begin {gather*} \frac {(a-b) \text {ArcTan}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 b^{3/2} f}-\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 396

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

Rule 3265

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 105, normalized size = 1.27 \begin {gather*} -\frac {\cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))}}{2 \sqrt {2} b f}+\frac {(a-b) \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{2 \sqrt {-b} b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])/(Sqrt[2]*b*f) + ((a - b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f
*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/(2*Sqrt[-b]*b*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(71)=142\).
time = 7.69, size = 186, normalized size = 2.24

method	result	size
default	\(-\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (2 b^{\frac {3}{2}} \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}-b^{2} \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )+b a \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )\right )}{4 b^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.57, size = 79, normalized size = 0.95 \begin {gather*} \frac {\frac {a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {\arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{b}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (71) = 142\).
time = 0.51, size = 438, normalized size = 5.28 \begin {gather*} \left [-\frac {8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) - {\left (a - b\right )} \sqrt {-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}\right )}{16 \, b^{2} f}, -\frac {{\left (a - b\right )} \sqrt {b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right ) + 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right )}{8 \, b^{2} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e) - (a - b)*sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(
a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b
^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 + 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)
*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sq
rt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)))/(b^2*f), -1/8*((a - b)*sqrt(b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(
a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 -
3*(a*b^2 + b^3)*cos(f*x + e)^3 + (a^2*b + 2*a*b^2 + b^3)*cos(f*x + e))) + 4*sqrt(-b*cos(f*x + e)^2 + a + b)*b
*cos(f*x + e))/(b^2*f)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.91, size = 91, normalized size = 1.10 \begin {gather*} -\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{2 \, b f} + \frac {{\left (a - b\right )} \log \left ({\left | \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} + \frac {\sqrt {-b f^{2}} \cos \left (f x + e\right )}{f} \right |}\right )}{2 \, \sqrt {-b} b {\left | f \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^3}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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