3.300 \(\int \tan ^4(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=169 \[ -\frac{\left (a^2+4 a b-8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 b^{3/2} f}+\frac{\tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 f}+\frac{(a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 b f}+\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f} \]

[Out]

(Sqrt[a - b]*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f - ((a^2 + 4*a*b - 8*b^2)*ArcTanh
[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*b^(3/2)*f) + ((a - 4*b)*Tan[e + f*x]*Sqrt[a + b*Tan[e
+ f*x]^2])/(8*b*f) + (Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(4*f)

________________________________________________________________________________________

Rubi [A]  time = 0.210355, antiderivative size = 169, 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, 478, 582, 523, 217, 206, 377, 203} \[ -\frac{\left (a^2+4 a b-8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 b^{3/2} f}+\frac{\tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 f}+\frac{(a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 b f}+\frac{\sqrt{a-b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^4*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 + 4*a*b - 8*b^2)*ArcTanh
[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*b^(3/2)*f) + ((a - 4*b)*Tan[e + f*x]*Sqrt[a + b*Tan[e
+ f*x]^2])/(8*b*f) + (Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(4*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 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 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 ^4(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \sqrt{a+b x^2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 f}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a+(-a+4 b) x^2\right )}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac{(a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 b f}+\frac{\tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 f}+\frac{\operatorname{Subst}\left (\int \frac{-a (a-4 b)+\left (-a^2-4 a b+8 b^2\right ) x^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 b f}\\ &=\frac{(a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 b f}+\frac{\tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 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{\left (a^2+4 a b-8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 b f}\\ &=\frac{(a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 b f}+\frac{\tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 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{\left (a^2+4 a b-8 b^2\right ) \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 )}{8 b 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{\left (a^2+4 a b-8 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 b^{3/2} f}+\frac{(a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 b f}+\frac{\tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 f}\\ \end{align*}

Mathematica [C]  time = 6.20682, size = 767, normalized size = 4.54 \[ \frac{\sqrt{\frac{a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b}{\cos (2 (e+f x))+1}} \left (\frac{\sec (e+f x) (a \sin (e+f x)-6 b \sin (e+f x))}{8 b}+\frac{1}{4} \tan (e+f x) \sec ^2(e+f x)\right )}{f}-\frac{-\frac{b \left (a^2-4 b^2\right ) \sin ^4(e+f x) \csc (2 (e+f x)) \sqrt{\frac{(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )}{a ((a-b) \cos (2 (e+f x))+a+b)}-\frac{4 b \left (4 b^2-4 a b\right ) \sqrt{\cos (2 (e+f x))+1} \sqrt{\frac{(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \left (\frac{\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )}{4 a \sqrt{\cos (2 (e+f x))+1} \sqrt{(a-b) \cos (2 (e+f x))+a+b}}-\frac{\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )}{2 (a-b) \sqrt{\cos (2 (e+f x))+1} \sqrt{(a-b) \cos (2 (e+f x))+a+b}}\right )}{\sqrt{(a-b) \cos (2 (e+f x))+a+b}}}{4 b f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^4*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-(-((b*(a^2 - 4*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]/Sqrt[2]]
, 1]*Sin[e + f*x]^4)/(a*(a + b + (a - b)*Cos[2*(e + f*x)]))) - (4*b*(-4*a*b + 4*b^2)*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)]*S
qrt[-((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)]])/(4*b*f) + (Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 +
Cos[2*(e + f*x)])]*((Sec[e + f*x]*(a*Sin[e + f*x] - 6*b*Sin[e + f*x]))/(8*b) + (Sec[e + f*x]^2*Tan[e + f*x])/
4))/f

________________________________________________________________________________________

Maple [B]  time = 0.03, size = 323, normalized size = 1.9 \begin{align*}{\frac{\tan \left ( fx+e \right ) }{4\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{a\tan \left ( fx+e \right ) }{8\,fb}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}-{\frac{{a}^{2}}{8\,f}\ln \left ( \sqrt{b}\tan \left ( fx+e \right ) +\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}} \right ){b}^{-{\frac{3}{2}}}}-{\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)^4,x)

[Out]

1/4/f*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2)/b-1/8/f*a/b*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)-1/8/f*a^2/b^(3/2)*ln
(b^(1/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))-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))

________________________________________________________________________________________

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 )^{4}\,{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)^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 7.59838, size = 1675, normalized size = 9.91 \begin{align*} \text{result too large to display} \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)^4,x, algorithm="fricas")

[Out]

[1/16*(8*sqrt(-a + b)*b^2*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 + 4*a*b - 8*b^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*(2*b^2*tan(f*x + e)^3 + (a*b - 4*b^2)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2
+ a))/(b^2*f), 1/16*(16*sqrt(a - b)*b^2*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) - (a^2
+ 4*a*b - 8*b^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*
(2*b^2*tan(f*x + e)^3 + (a*b - 4*b^2)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/8*(4*sqrt(-a + b)*b
^2*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 + 4*a*b - 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + (2*b^2
*tan(f*x + e)^3 + (a*b - 4*b^2)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/8*(8*sqrt(a - b)*b^2*arct
an(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + (a^2 + 4*a*b - 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(
f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + (2*b^2*tan(f*x + e)^3 + (a*b - 4*b^2)*tan(f*x + e))*sqrt(b*tan(f*
x + e)^2 + a))/(b^2*f)]

________________________________________________________________________________________

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

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*tan(e + f*x)**4, x)

________________________________________________________________________________________

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 )^{4}\,{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)^4,x, algorithm="giac")

[Out]

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