3.352 \(\int \frac{\tan ^6(e+f x)}{(a+b \tan ^2(e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=171 \[ -\frac{a (a-2 b) \tan (e+f x)}{b^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac{a \tan ^3(e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}} \]
[Out]
-(ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/((a - b)^(5/2)*f)) + ArcTanh[(Sqrt[b]*Tan[e +
f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(b^(5/2)*f) - (a*Tan[e + f*x]^3)/(3*(a - b)*b*f*(a + b*Tan[e + f*x]^2)^(3/2)
) - (a*(a - 2*b)*Tan[e + f*x])/((a - b)^2*b^2*f*Sqrt[a + b*Tan[e + f*x]^2])
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Rubi [A] time = 0.266733, antiderivative size = 171, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used =
{3670, 470, 578, 523, 217, 206, 377, 203} \[ -\frac{a (a-2 b) \tan (e+f x)}{b^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac{a \tan ^3(e+f x)}{3 b f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^6/(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)) + ArcTanh[(Sqrt[b]*Tan[e +
f*x])/Sqrt[a + b*Tan[e + f*x]^2]]/(b^(5/2)*f) - (a*Tan[e + f*x]^3)/(3*(a - b)*b*f*(a + b*Tan[e + f*x]^2)^(3/2)
) - (a*(a - 2*b)*Tan[e + f*x])/((a - b)^2*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 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 578
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[(g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c -
a*d)*(p + 1)), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]
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 \frac{\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan ^3(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{x^2 \left (3 a+3 (a-b) x^2\right )}{\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 ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b) \tan (e+f x)}{(a-b)^2 b^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{-3 a (a-2 b)-3 (a-b)^2 x^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 (a-b)^2 b^2 f}\\ &=-\frac{a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b) \tan (e+f x)}{(a-b)^2 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{\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{b^2 f}\\ &=-\frac{a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b) \tan (e+f x)}{(a-b)^2 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{\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 )}{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{\tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{b^{5/2} f}-\frac{a \tan ^3(e+f x)}{3 (a-b) b f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac{a (a-2 b) \tan (e+f x)}{(a-b)^2 b^2 f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 4.44009, size = 295, normalized size = 1.73 \[ -\frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (a^2 (a-b) \sin (2 (e+f x)) ((3 a-7 b) ((a-b) \cos (2 (e+f x))+a+b)+2 a b)-\frac{3 a^2 b \sin ^2(e+f x) \sin (2 (e+f x)) \left (\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}\right )^{3/2} \left (\left (a^2-3 a b+2 b^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )+b^2 \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{\sqrt{2}}\right )}{3 \sqrt{2} a b^2 f (a-b)^3 ((a-b) \cos (2 (e+f x))+a+b)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(5/2),x]
[Out]
-(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(a^2*(a - b)*(2*a*b + (3*a - 7*b)*(a + b + (a - b)*C
os[2*(e + f*x)]))*Sin[2*(e + f*x)] - (3*a^2*b*(((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)^(3/2)*((
a^2 - 3*a*b + 2*b^2)*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]
+ b^2*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1
])*Sin[e + f*x]^2*Sin[2*(e + f*x)])/Sqrt[2]))/(3*Sqrt[2]*a*(a - b)^3*b^2*f*(a + b + (a - b)*Cos[2*(e + f*x)])^
2)
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Maple [B] time = 0.047, size = 382, normalized size = 2.2 \begin{align*} -{\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{\tan \left ( fx+e \right ) }{f{b}^{2}}{\frac{1}{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{1}{f}\ln \left ( \sqrt{b}\tan \left ( fx+e \right ) +\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}} \right ){b}^{-{\frac{5}{2}}}}+{\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)^6/(a+b*tan(f*x+e)^2)^(5/2),x)
[Out]
-1/3/f*tan(f*x+e)^3/b/(a+b*tan(f*x+e)^2)^(3/2)-1/f/b^2*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)+1/f/b^(5/2)*ln(b^(1
/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))+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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [B] time = 18.9618, size = 3800, normalized size = 22.22 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
[1/6*(3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b
- 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(f*x + e)^2)*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) - 3*(b^5*tan(f*x + e)^4 + 2*a*b^4*tan(f*x + e)^2 + a^2*b^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*
((4*a^3*b^2 - 11*a^2*b^3 + 7*a*b^4)*tan(f*x + e)^3 + 3*(a^4*b - 3*a^3*b^2 + 2*a^2*b^3)*tan(f*x + e))*sqrt(b*ta
n(f*x + e)^2 + a))/((a^3*b^5 - 3*a^2*b^6 + 3*a*b^7 - b^8)*f*tan(f*x + e)^4 + 2*(a^4*b^4 - 3*a^3*b^5 + 3*a^2*b^
6 - a*b^7)*f*tan(f*x + e)^2 + (a^5*b^3 - 3*a^4*b^4 + 3*a^3*b^5 - a^2*b^6)*f), -1/6*(6*(a^5 - 3*a^4*b + 3*a^3*b
^2 - a^2*b^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4
)*tan(f*x + e)^2)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + 3*(b^5*tan(f*x + e)^
4 + 2*a*b^4*tan(f*x + e)^2 + a^2*b^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*((4*a^3*b^2 - 11*a^2*b^3 + 7*a*b^4)*tan(f*x + e)^3
+ 3*(a^4*b - 3*a^3*b^2 + 2*a^2*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^3*b^5 - 3*a^2*b^6 + 3*a*b^7
- b^8)*f*tan(f*x + e)^4 + 2*(a^4*b^4 - 3*a^3*b^5 + 3*a^2*b^6 - a*b^7)*f*tan(f*x + e)^2 + (a^5*b^3 - 3*a^4*b^4
+ 3*a^3*b^5 - a^2*b^6)*f), -1/6*(6*(b^5*tan(f*x + e)^4 + 2*a*b^4*tan(f*x + e)^2 + a^2*b^3)*sqrt(a - b)*arctan
(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - 3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3*b^2 -
3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(f*x + e)^2)*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*((4*a^3*b^2 - 11*a^2*b^3
+ 7*a*b^4)*tan(f*x + e)^3 + 3*(a^4*b - 3*a^3*b^2 + 2*a^2*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/((a^3*
b^5 - 3*a^2*b^6 + 3*a*b^7 - b^8)*f*tan(f*x + e)^4 + 2*(a^4*b^4 - 3*a^3*b^5 + 3*a^2*b^6 - a*b^7)*f*tan(f*x + e)
^2 + (a^5*b^3 - 3*a^4*b^4 + 3*a^3*b^5 - a^2*b^6)*f), -1/3*(3*(b^5*tan(f*x + e)^4 + 2*a*b^4*tan(f*x + e)^2 + a^
2*b^3)*sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + 3*(a^5 - 3*a^4*b + 3*a^3*b
^2 - a^2*b^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4
)*tan(f*x + e)^2)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + ((4*a^3*b^2 - 11*a^2
*b^3 + 7*a*b^4)*tan(f*x + e)^3 + 3*(a^4*b - 3*a^3*b^2 + 2*a^2*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(
(a^3*b^5 - 3*a^2*b^6 + 3*a*b^7 - b^8)*f*tan(f*x + e)^4 + 2*(a^4*b^4 - 3*a^3*b^5 + 3*a^2*b^6 - a*b^7)*f*tan(f*x
+ e)^2 + (a^5*b^3 - 3*a^4*b^4 + 3*a^3*b^5 - a^2*b^6)*f)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{6}{\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)**6/(a+b*tan(f*x+e)**2)**(5/2),x)
[Out]
Integral(tan(e + f*x)**6/(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 )^{6}}{{\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)^6/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="giac")
[Out]
integrate(tan(f*x + e)^6/(b*tan(f*x + e)^2 + a)^(5/2), x)