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

Optimal. Leaf size=123 \[ -\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{\tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}+\frac{(a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 \sqrt{b} f} \]

[Out]

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

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Rubi [A]  time = 0.133592, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3670, 478, 523, 217, 206, 377, 203} \[ -\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}+\frac{\tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{2 f}+\frac{(a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 \sqrt{b} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 478

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

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, 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])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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

Mathematica [C]  time = 5.86573, size = 251, normalized size = 2.04 \[ \frac{\tan (e+f x) \left (-\sqrt{2} a \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )+\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)+2 \sqrt{2} a \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{2 \sqrt{2} f \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-(Sqrt[2]*a*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcSin[Sqrt[((a + b + (a -
b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]) + 2*Sqrt[2]*a*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*
Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]
/Sqrt[2]], 1] + (a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2)*Tan[e + f*x])/(2*Sqrt[2]*f*Sqrt[(a + b + (a
- b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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Maple [B]  time = 0.028, size = 230, normalized size = 1.9 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{2\,f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{a}{2\,f}\ln \left ( \sqrt{b}\tan \left ( fx+e \right ) +\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}} \right ){\frac{1}{\sqrt{b}}}}-{\frac{1}{f}\sqrt{b}\ln \left ( \sqrt{b}\tan \left ( fx+e \right ) +\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}} \right ) }+{\frac{1}{fb \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) }-{\frac{a}{f{b}^{2} \left ( a-b \right ) }\sqrt{{b}^{4} \left ( a-b \right ) }\arctan \left ({ \left ( a-b \right ){b}^{2}\tan \left ( fx+e \right ){\frac{1}{\sqrt{{b}^{4} \left ( a-b \right ) }}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*tan(f*x + e)^2, x)

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Fricas [A]  time = 3.45395, size = 1388, normalized size = 11.28 \begin{align*} \left [-\frac{{\left (a - 2 \, b\right )} \sqrt{b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{b} \tan \left (f x + e\right ) + a\right ) - 2 \, \sqrt{-a + b} b \log \left (-\frac{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, b f}, -\frac{4 \, \sqrt{a - b} b \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\right ) +{\left (a - 2 \, b\right )} \sqrt{b} \log \left (2 \, b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{b} \tan \left (f x + e\right ) + a\right ) - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{4 \, b f}, -\frac{{\left (a - 2 \, b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-b}}{b \tan \left (f x + e\right )}\right ) - \sqrt{-a + b} b \log \left (-\frac{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b} \tan \left (f x + e\right ) - a}{\tan \left (f x + e\right )^{2} + 1}\right ) - \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{2 \, b f}, -\frac{2 \, \sqrt{a - b} b \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{a - b} \tan \left (f x + e\right )}\right ) +{\left (a - 2 \, b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-b}}{b \tan \left (f x + e\right )}\right ) - \sqrt{b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right )}{2 \, b f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((a - 2*b)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(b)*tan(f*x + e) + a) - 2*s
qrt(-a + b)*b*log(-((a - 2*b)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(ta
n(f*x + e)^2 + 1)) - 2*sqrt(b*tan(f*x + e)^2 + a)*b*tan(f*x + e))/(b*f), -1/4*(4*sqrt(a - b)*b*arctan(-sqrt(b*
tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + (a - 2*b)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x
+ e)^2 + a)*sqrt(b)*tan(f*x + e) + a) - 2*sqrt(b*tan(f*x + e)^2 + a)*b*tan(f*x + e))/(b*f), -1/2*((a - 2*b)*sq
rt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) - sqrt(-a + b)*b*log(-((a - 2*b)*tan(f*x +
e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - sqrt(b*tan(f*x + e
)^2 + a)*b*tan(f*x + e))/(b*f), -1/2*(2*sqrt(a - b)*b*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x
+ e))) + (a - 2*b)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) - sqrt(b*tan(f*x + e)
^2 + a)*b*tan(f*x + e))/(b*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 )}} \tan ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*tan(f*x + e)^2, x)