3.123 \(\int \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)} \, dx\)

Optimal. Leaf size=78 \[ -\frac{\cos (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{2 f}-\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt{b} f} \]

[Out]

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

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Rubi [A]  time = 0.0609011, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3186, 195, 217, 203} \[ -\frac{\cos (e+f x) \sqrt{a-b \cos ^2(e+f x)+b}}{2 f}-\frac{(a+b) \tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt{b} f} \]

Antiderivative was successfully verified.

[In]

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

Rule 3186

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]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

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 203

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

Rubi steps

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

Mathematica [A]  time = 0.253022, size = 93, normalized size = 1.19 \[ -\frac{\sqrt{2} \cos (e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b}+\frac{2 (a+b) \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )}{\sqrt{-b}}}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[2]*Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]] + (2*(a + b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqr
t[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[-b])/(4*f)

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Maple [B]  time = 1.282, size = 182, normalized size = 2.3 \begin{align*}{\frac{1}{4\,f\cos \left ( fx+e \right ) }\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) } \left ( b\arctan \left ({\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{2}{\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ) +a\arctan \left ({\frac{-2\,b \left ( \cos \left ( fx+e \right ) \right ) ^{2}+a+b}{2}{\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ) -2\,\sqrt{b}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ){\frac{1}{\sqrt{b}}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.89486, size = 1069, normalized size = 13.71 \begin{align*} \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 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 f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)*(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*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*co
s(f*x + e))/(b*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sin(e + f*x)**2)*sin(e + f*x), x)

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Giac [A]  time = 1.73559, size = 112, normalized size = 1.44 \begin{align*} -\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{2 \, 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}{\left | f \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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