3.354 \(\int \frac{\tan ^2(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=128 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}+\frac{(2 a+b) \tan (e+f x)}{3 a f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tan (e+f x)}{3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) + Tan[e + f*x]/(3*(a - b)*f
*(a + b*Tan[e + f*x]^2)^(3/2)) + ((2*a + b)*Tan[e + f*x])/(3*a*(a - b)^2*f*Sqrt[a + b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.152694, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 471, 527, 12, 377, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}+\frac{(2 a+b) \tan (e+f x)}{3 a f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tan (e+f x)}{3 f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) + Tan[e + f*x]/(3*(a - b)*f
*(a + b*Tan[e + f*x]^2)^(3/2)) + ((2*a + b)*Tan[e + f*x])/(3*a*(a - b)^2*f*Sqrt[a + b*Tan[e + f*x]^2])

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 471

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 + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 \frac{\tan ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{1-2 x^2}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 (a-b) f}\\ &=\frac{\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(2 a+b) \tan (e+f x)}{3 a (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{3 a}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b)^2 f}\\ &=\frac{\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(2 a+b) \tan (e+f x)}{3 a (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}\\ &=\frac{\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(2 a+b) \tan (e+f x)}{3 a (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\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 )}{(a-b)^2 f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{(a-b)^{5/2} f}+\frac{\tan (e+f x)}{3 (a-b) f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(2 a+b) \tan (e+f x)}{3 a (a-b)^2 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 7.73209, size = 365, normalized size = 2.85 \[ \frac{\cos ^4(e+f x) \cot (e+f x) \left (12 (a-b)^3 \tan ^6(e+f x) \left (a+b \tan ^2(e+f x)\right ) \text{Hypergeometric2F1}\left (2,2,\frac{9}{2},\frac{(a-b) \sin ^2(e+f x)}{a}\right ) \sqrt{\frac{\sin ^2(e+f x) \cos ^2(e+f x) \left (a^2+a b \left (\tan ^2(e+f x)-1\right )-b^2 \tan ^2(e+f x)\right )}{a^2}}+35 a \sec ^2(e+f x) \left (5 a+2 b \tan ^2(e+f x)\right ) \left (a \sec ^2(e+f x) \left (a \left (\tan ^2(e+f x)-3\right )-4 b \tan ^2(e+f x)\right ) \sqrt{\frac{\sin ^2(e+f x) \cos ^2(e+f x) \left (a^2+a b \left (\tan ^2(e+f x)-1\right )-b^2 \tan ^2(e+f x)\right )}{a^2}}+3 \sin ^{-1}\left (\sqrt{\frac{(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a+b \tan ^2(e+f x)\right )^2\right )\right )}{315 a^4 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)} \left (\frac{b \tan ^2(e+f x)}{a}+1\right ) \sqrt{\frac{(a-b) \sin ^2(e+f x) \cos ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )}{a^2}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[e + f*x]^2/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

(Cos[e + f*x]^4*Cot[e + f*x]*(12*(a - b)^3*Hypergeometric2F1[2, 2, 9/2, ((a - b)*Sin[e + f*x]^2)/a]*Tan[e + f*
x]^6*(a + b*Tan[e + f*x]^2)*Sqrt[(Cos[e + f*x]^2*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e +
f*x]^2)))/a^2] + 35*a*Sec[e + f*x]^2*(5*a + 2*b*Tan[e + f*x]^2)*(3*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*(a
+ b*Tan[e + f*x]^2)^2 + a*Sec[e + f*x]^2*(-4*b*Tan[e + f*x]^2 + a*(-3 + Tan[e + f*x]^2))*Sqrt[(Cos[e + f*x]^2
*Sin[e + f*x]^2*(a^2 - b^2*Tan[e + f*x]^2 + a*b*(-1 + Tan[e + f*x]^2)))/a^2])))/(315*a^4*(a - b)^2*f*Sqrt[a +
b*Tan[e + f*x]^2]*Sqrt[((a - b)*Cos[e + f*x]^2*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2))/a^2]*(1 + (b*Tan[e + f*x
]^2)/a))

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Maple [B]  time = 0.044, size = 232, normalized size = 1.8 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{3\,fa} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,\tan \left ( fx+e \right ) }{3\,f{a}^{2}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{b\tan \left ( fx+e \right ) }{f \left ( a-b \right ) ^{2}a}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{1}{f \left ( a-b \right ) ^{3}{b}^{2}}\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{b\tan \left ( fx+e \right ) }{3\,a \left ( a-b \right ) f} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,b\tan \left ( fx+e \right ) }{3\,f \left ( a-b \right ){a}^{2}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x)

[Out]

1/3/f*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(3/2)+2/3/f/a^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)+1/f/(a-b)^2*b*tan(f*
x+e)/a/(a+b*tan(f*x+e)^2)^(1/2)-1/f/(a-b)^3*(b^4*(a-b))^(1/2)/b^2*arctan(b^2*(a-b)/(b^4*(a-b))^(1/2)/(a+b*tan(
f*x+e)^2)^(1/2)*tan(f*x+e))+1/3*b*tan(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)^(3/2)+2/3/f/(a-b)*b/a^2*tan(f*x+e)/(
a+b*tan(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(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14815, size = 1176, normalized size = 9.19 \begin{align*} \left [-\frac{3 \,{\left (a b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} b \tan \left (f x + e\right )^{2} + a^{3}\right )} \sqrt{-a + 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 \,{\left ({\left (2 \, a^{2} b - a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \,{\left (a^{3} - a^{2} b\right )} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{6 \,{\left ({\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} f\right )}}, -\frac{3 \,{\left (a b^{2} \tan \left (f x + e\right )^{4} + 2 \, a^{2} b \tan \left (f x + e\right )^{2} + a^{3}\right )} \sqrt{a - 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 ({\left (2 \, a^{2} b - a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \,{\left (a^{3} - a^{2} b\right )} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{3 \,{\left ({\left (a^{4} b^{2} - 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} - a b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{5} b - 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} - a^{2} b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{6} - 3 \, a^{5} b + 3 \, a^{4} b^{2} - a^{3} b^{3}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*(3*(a*b^2*tan(f*x + e)^4 + 2*a^2*b*tan(f*x + e)^2 + a^3)*sqrt(-a + 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)) - 2*((2*a^2*b - a*b^2 - b^3)*
tan(f*x + e)^3 + 3*(a^3 - a^2*b)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 -
a*b^5)*f*tan(f*x + e)^4 + 2*(a^5*b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4)*f*tan(f*x + e)^2 + (a^6 - 3*a^5*b + 3*a
^4*b^2 - a^3*b^3)*f), -1/3*(3*(a*b^2*tan(f*x + e)^4 + 2*a^2*b*tan(f*x + e)^2 + a^3)*sqrt(a - b)*arctan(-sqrt(b
*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - ((2*a^2*b - a*b^2 - b^3)*tan(f*x + e)^3 + 3*(a^3 - a^2*b)*t
an(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^4*b^2 - 3*a^3*b^3 + 3*a^2*b^4 - a*b^5)*f*tan(f*x + e)^4 + 2*(a^5*
b - 3*a^4*b^2 + 3*a^3*b^3 - a^2*b^4)*f*tan(f*x + e)^2 + (a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral(tan(e + f*x)**2/(a + b*tan(e + f*x)**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^2/(b*tan(f*x + e)^2 + a)^(5/2), x)