3.339 \(\int \frac {\tan ^6(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=182 \[ -\frac {(3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}+\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b^2 f (a-b)}-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}}-\frac {a \tan ^3(e+f x)}{b f (a-b) \sqrt {a+b \tan ^2(e+f x)}} \]

[Out]

-arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(3/2)/f-1/2*(3*a+2*b)*arctanh(b^(1/2)*tan(f*x+e
)/(a+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f+1/2*(3*a-b)*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/(a-b)/b^2/f-a*tan(f*x+e)
^3/(a-b)/b/f/(a+b*tan(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3670, 470, 582, 523, 217, 206, 377, 203} \[ \frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b^2 f (a-b)}-\frac {(3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {a \tan ^3(e+f x)}{b f (a-b) \sqrt {a+b \tan ^2(e+f x)}}-\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f (a-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

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

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])

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 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 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 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 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 582

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

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]))

Rubi steps

\begin {align*} \int \frac {\tan ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a \tan ^3(e+f x)}{(a-b) b f \sqrt {a+b \tan ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a+(3 a-b) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{(a-b) b f}\\ &=-\frac {a \tan ^3(e+f x)}{(a-b) b f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) b^2 f}-\frac {\operatorname {Subst}\left (\int \frac {a (3 a-b)+(a-b) (3 a+2 b) x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 (a-b) b^2 f}\\ &=-\frac {a \tan ^3(e+f x)}{(a-b) b f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) b^2 f}-\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) f}-\frac {(3 a+2 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 b^2 f}\\ &=-\frac {a \tan ^3(e+f x)}{(a-b) b f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) b^2 f}-\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) f}-\frac {(3 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 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)^{3/2} f}-\frac {(3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {a \tan ^3(e+f x)}{(a-b) b f \sqrt {a+b \tan ^2(e+f x)}}+\frac {(3 a-b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 (a-b) b^2 f}\\ \end {align*}

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Mathematica [C]  time = 6.44, size = 787, normalized size = 4.32 \[ \frac {\sqrt {\frac {a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b}{\cos (2 (e+f x))+1}} \left (\frac {\tan (e+f x)}{2 b^2}-\frac {a^2 \sin (2 (e+f x))}{b^2 (a-b) (a (-\cos (2 (e+f x)))-a+b \cos (2 (e+f x))-b)}\right )}{f}-\frac {-\frac {b \left (3 a^2-a b-b^2\right ) \sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {\frac {(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{a ((a-b) \cos (2 (e+f x))+a+b)}-\frac {4 b^3 \sqrt {\cos (2 (e+f x))+1} \sqrt {\frac {(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \left (\frac {\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{4 a \sqrt {\cos (2 (e+f x))+1} \sqrt {(a-b) \cos (2 (e+f x))+a+b}}-\frac {\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{2 (a-b) \sqrt {\cos (2 (e+f x))+1} \sqrt {(a-b) \cos (2 (e+f x))+a+b}}\right )}{\sqrt {(a-b) \cos (2 (e+f x))+a+b}}}{b^2 f (a-b)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[e + f*x]^6/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-((-((b*(3*a^2 - a*b - b^2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*Sqrt[-((a*Cot[e +
f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[
e + f*x]^2)/b]*Csc[2*(e + f*x)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/S
qrt[2]], 1]*Sin[e + f*x]^4)/(a*(a + b + (a - b)*Cos[2*(e + f*x)]))) - (4*b^3*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[(
a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a*(1 + Cos[2
*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*(e + f*x)]*
EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*Sin[e + f*x]^4)/(4*a
*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]) - (Sqrt[-((a*Cot[e + f*x]^2)/b)]*Sqrt[-((a
*(1 + Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b)]*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*Csc[2*
(e + f*x)]*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]^4)/(2*(a - b)*Sqrt[1 + Cos[2*(e + f*x)]]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])))/Sqrt[a +
b + (a - b)*Cos[2*(e + f*x)]])/((a - b)*b^2*f)) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1
+ Cos[2*(e + f*x)])]*(-((a^2*Sin[2*(e + f*x)])/((a - b)*b^2*(-a - b - a*Cos[2*(e + f*x)] + b*Cos[2*(e + f*x)])
)) + Tan[e + f*x]/(2*b^2)))/f

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fricas [A]  time = 2.34, size = 1207, normalized size = 6.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (f x + e\right )^{6}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^6/(b*tan(f*x + e)^2 + a)^(3/2), x)

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maple [A]  time = 0.39, size = 286, normalized size = 1.57 \[ \frac {\tan ^{3}\left (f x +e \right )}{2 f b \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {3 a \tan \left (f x +e \right )}{2 f \,b^{2} \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}-\frac {3 a \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 f \,b^{\frac {5}{2}}}+\frac {\tan \left (f x +e \right )}{f b \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}-\frac {\ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{f \,b^{\frac {3}{2}}}+\frac {\tan \left (f x +e \right )}{f a \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}+\frac {b \tan \left (f x +e \right )}{a \left (a -b \right ) f \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \left (a -b \right )^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

1/2/f*tan(f*x+e)^3/b/(a+b*tan(f*x+e)^2)^(1/2)+3/2/f/b^2*a*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2)-3/2/f/b^(5/2)*a*
ln(tan(f*x+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))+1/f*tan(f*x+e)/b/(a+b*tan(f*x+e)^2)^(1/2)-1/f/b^(3/2)*ln(tan(f
*x+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))+1/f*tan(f*x+e)/a/(a+b*tan(f*x+e)^2)^(1/2)+b*tan(f*x+e)/a/(a-b)/f/(a+b*
tan(f*x+e)^2)^(1/2)-1/f/(a-b)^2*(b^4*(a-b))^(1/2)/b^2*arctan((a-b)*b^2/(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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^6/(a + b*tan(e + f*x)^2)^(3/2),x)

[Out]

int(tan(e + f*x)^6/(a + b*tan(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**6/(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Integral(tan(e + f*x)**6/(a + b*tan(e + f*x)**2)**(3/2), x)

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