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

Optimal. Leaf size=58 \[ \frac {\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f}-\frac {\sqrt {a+b \sin ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 58, 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 = {3194, 50, 63, 208} \[ \frac {\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f}-\frac {\sqrt {a+b \sin ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 3194

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 60, normalized size = 1.03 \[ \frac {\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a-b \cos ^2(e+f x)+b}}{\sqrt {a+b}}\right )-\sqrt {a-b \cos ^2(e+f x)+b}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 145, normalized size = 2.50 \[ \left [\frac {\sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{2 \, f}, -\frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.32, size = 329, normalized size = 5.67 \[ -\frac {2 \, {\left (\frac {{\left (a + b\right )} \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} - \sqrt {a}}{2 \, \sqrt {-a - b}}\right )}{\sqrt {-a - b}} - \frac {2 \, {\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} b - \sqrt {a} b\right )}}{{\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )} \sqrt {a} + a + 4 \, b}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((a + b)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/
2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/sqrt(-a - b) - 2*((sqrt(a)*tan(1/2*f*x + 1/2
*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*b - sqrt
(a)*b)/((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan
(1/2*f*x + 1/2*e)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*
f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) + a + 4*b))/f

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maple [B]  time = 4.14, size = 134, normalized size = 2.31 \[ \frac {\sqrt {a +b}\, \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right )}{2 f}+\frac {\sqrt {a +b}\, \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right )}{2 f}-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 122, normalized size = 2.10 \[ -\frac {\sqrt {a + b} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right ) - \sqrt {a + b} \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right ) + 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(a + b)*arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1))) -
sqrt(a + b)*arcsinh(-b*sin(f*x + e)/(sqrt(a*b)*(sin(f*x + e) - 1)) - a/(sqrt(a*b)*(sin(f*x + e) - 1))) + 2*sqr
t(b*sin(f*x + e)^2 + a))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {tan}\left (e+f\,x\right )\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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