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]

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

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Rubi [A]  time = 0.06, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.56, 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]

-1/4*((-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)])
/(f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])

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fricas [A]  time = 0.70, size = 203, normalized size = 2.82 \[ \left [-\frac {2 \, \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, -\frac {\sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) + \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{f}\right ] \]

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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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)*(a+b*tan(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)abs(f)*(-sqrt(a*f^
2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)*sign(cos(f*x+exp(1)))*sign(f)/f^2-b*sign(cos(f*x+exp(
1)))*sign(f)*atan(sqrt(a*f^2*(-cos(f*x+exp(1))/f)^2-b*f^2*(-cos(f*x+exp(1))/f)^2+b)/sqrt(-b))/sqrt(-b)/f^2-(-b
*atan(sqrt(b)/sqrt(-b))-sqrt(-b)*sqrt(b))/f^2/sqrt(-b)*sign(cos(f*x+exp(1)))*sign(f))

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maple [B]  time = 0.39, size = 144, normalized size = 2.00 \[ \frac {\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) \left (\sqrt {b}\, \ln \left (\frac {2 \sqrt {b}\, \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}+2 b}{\cos \left (f x +e \right )}\right )-\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}\right )}{f \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}} \]

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 [A]  time = 0.45, size = 97, normalized size = 1.35 \[ -\frac {2 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{2 \, f} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (e+f\,x\right )\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx \]

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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