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

Optimal. Leaf size=66 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]  time = 0.0533864, antiderivative size = 66, 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 = {4134, 277, 217, 206} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]
^2])/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 277

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

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 206

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

Rubi steps

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

Mathematica [A]  time = 0.150015, size = 98, normalized size = 1.48 \[ \frac{\sqrt{2} \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)} \left (\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)+b}}{\sqrt{b}}\right )-\sqrt{a \cos ^2(e+f x)+b}\right )}{f \sqrt{a \cos (2 (e+f x))+a+2 b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.077, size = 93, normalized size = 1.4 \begin{align*} -{\frac{1}{fa\sec \left ( fx+e \right ) } \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{b\sec \left ( fx+e \right ) }{fa}\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}}}+{\frac{1}{f}\sqrt{b}\ln \left ( \sec \left ( fx+e \right ) \sqrt{b}+\sqrt{a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.07868, size = 468, normalized size = 7.09 \begin{align*} \left [-\frac{2 \, \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt{b} \log \left (\frac{a \cos \left (f x + e\right )^{2} + 2 \, \sqrt{b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, -\frac{\sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) + \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2))/f, -(sqrt(-b)*arctan(sqrt(-b)
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos
(f*x + e))/f]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec ^{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*sec(f*x+e)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.25855, size = 78, normalized size = 1.18 \begin{align*} -\frac{{\left (\frac{b \arctan \left (\frac{\sqrt{a \cos \left (f x + e\right )^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + \sqrt{a \cos \left (f x + e\right )^{2} + b}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*arctan(sqrt(a*cos(f*x + e)^2 + b)/sqrt(-b))/sqrt(-b) + sqrt(a*cos(f*x + e)^2 + b))*sgn(cos(f*x + e))/f