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

Optimal. Leaf size=72 \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f} \]

[Out]

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

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Rubi [A]  time = 0.0573515, antiderivative size = 72, 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 = {3664, 277, 217, 206} \[ \frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{f}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3664

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

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

Mathematica [A]  time = 0.544072, size = 140, normalized size = 1.94 \[ -\frac{\sin (2 (e+f x)) \csc (e+f x) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt{2} \sqrt{(a-b) \cos (2 (e+f x))+a+b}-2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{(a-b) \cos (2 (e+f x))+a+b}}{\sqrt{2} \sqrt{b}}\right )\right )}{4 f \sqrt{(a-b) \cos (2 (e+f x))+a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-2*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])] + Sqrt[2]*Sqrt[a + b + (a - b)
*Cos[2*(e + f*x)]])*Csc[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*Sin[2*(e + f*x)])/(4*
f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])

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Maple [B]  time = 0.069, size = 144, normalized size = 2. \begin{align*}{\frac{\cos \left ( fx+e \right ) }{f}\sqrt{{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}} \left ( \sqrt{b}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}+b}{\cos \left ( fx+e \right ) }} \right ) -\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b} \right ){\frac{1}{\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*cos(f*x+e)*(b^(1/2)*ln(2*(b^(1/2)*(a*cos(f*x+e)^2-c
os(f*x+e)^2*b+b)^(1/2)+b)/cos(f*x+e))-(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^(1/2))/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+
b)^(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*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.39196, size = 510, normalized size = 7.08 \begin{align*} \left [-\frac{2 \, \sqrt{\frac{{\left (a - b\right )} \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{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt{b} \sqrt{\frac{{\left (a - b\right )} \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{{\left (a - b\right )} \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{{\left (a - b\right )} \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*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(2*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b)*log(-((a - b)*cos(f*x + e)^2
+ 2*sqrt(b)*sqrt(((a - b)*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 - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + sqrt(((a - b)*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 \tan ^{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*tan(f*x+e)**2)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.61938, size = 176, normalized size = 2.44 \begin{align*} -{\left (\frac{{\left (\frac{b \arctan \left (\frac{\sqrt{a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + \sqrt{a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b}\right )} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{f^{2}} - \frac{{\left (b \arctan \left (\frac{\sqrt{b}}{\sqrt{-b}}\right ) + \sqrt{-b} \sqrt{b}\right )} \mathrm{sgn}\left (f\right ) \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{\sqrt{-b} f^{2}}\right )}{\left | f \right |} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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