3.353 \(\int \frac{\tan ^4(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=131 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}+\frac{(a-4 b) \tan (e+f x)}{3 b f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{a \tan (e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

[Out]

ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f) - (a*Tan[e + f*x])/(3*(a - b)*
b*f*(a + b*Tan[e + f*x]^2)^(3/2)) + ((a - 4*b)*Tan[e + f*x])/(3*(a - b)^2*b*f*Sqrt[a + b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.158553, antiderivative size = 131, 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, 470, 527, 12, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}}+\frac{(a-4 b) \tan (e+f x)}{3 b f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{a \tan (e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^4/(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) - (a*Tan[e + f*x])/(3*(a - b)*
b*f*(a + b*Tan[e + f*x]^2)^(3/2)) + ((a - 4*b)*Tan[e + f*x])/(3*(a - b)^2*b*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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 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] && GtQ[m - n + 1, n] && 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 ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{a+(a-3 b) 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) b f}\\ &=-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a b}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a (a-b)^2 b f}\\ &=-\frac{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b 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{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b 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{a \tan (e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{(a-4 b) \tan (e+f x)}{3 (a-b)^2 b f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 6.03819, size = 260, normalized size = 1.98 \[ \frac{\tan ^5(e+f x) \left (\frac{b \tan ^2(e+f x)}{a}+1\right ) \left (-\frac{\left (\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)\right )^2}{3 \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^2}-\frac{\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)}{\frac{b \tan ^2(e+f x)}{a}+1}+\frac{\sqrt{\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)} \tanh ^{-1}\left (\frac{\sqrt{\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)}}{\sqrt{\frac{b \tan ^2(e+f x)}{a}+1}}\right )}{\sqrt{\frac{b \tan ^2(e+f x)}{a}+1}}\right )}{a^2 f \sqrt{a+b \tan ^2(e+f x)} \left (\frac{b \tan ^2(e+f x)}{a}-\tan ^2(e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^4/(a + b*Tan[e + f*x]^2)^(5/2),x]

[Out]

(Tan[e + f*x]^5*(1 + (b*Tan[e + f*x]^2)/a)*((ArcTanh[Sqrt[-Tan[e + f*x]^2 + (b*Tan[e + f*x]^2)/a]/Sqrt[1 + (b*
Tan[e + f*x]^2)/a]]*Sqrt[-Tan[e + f*x]^2 + (b*Tan[e + f*x]^2)/a])/Sqrt[1 + (b*Tan[e + f*x]^2)/a] - (-Tan[e + f
*x]^2 + (b*Tan[e + f*x]^2)/a)/(1 + (b*Tan[e + f*x]^2)/a) - (-Tan[e + f*x]^2 + (b*Tan[e + f*x]^2)/a)^2/(3*(1 +
(b*Tan[e + f*x]^2)/a)^2)))/(a^2*f*Sqrt[a + b*Tan[e + f*x]^2]*(-Tan[e + f*x]^2 + (b*Tan[e + f*x]^2)/a)^3)

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Maple [B]  time = 0.046, size = 291, normalized size = 2.2 \begin{align*} -{\frac{\tan \left ( fx+e \right ) }{3\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{\tan \left ( fx+e \right ) }{3\,fab}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}-{\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)^4/(a+b*tan(f*x+e)^2)^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.19079, size = 1137, normalized size = 8.68 \begin{align*} \left [-\frac{3 \,{\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\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 (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \,{\left (a^{2} - a b\right )} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{6 \,{\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}, \frac{3 \,{\left (b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}\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 (a^{2} - 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - 3 \,{\left (a^{2} - a b\right )} \tan \left (f x + e\right )\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{3 \,{\left ({\left (a^{3} b^{2} - 3 \, a^{2} b^{3} + 3 \, a b^{4} - b^{5}\right )} f \tan \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} - a b^{4}\right )} f \tan \left (f x + e\right )^{2} +{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\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)**4/(a+b*tan(f*x+e)**2)**(5/2),x)

[Out]

Integral(tan(e + f*x)**4/(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 )^{4}}{{\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)^4/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^4/(b*tan(f*x + e)^2 + a)^(5/2), x)