3.313 \(\int \tan ^4(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=224 \[ \frac{\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{16 b f}-\frac{\left (6 a^2 b+a^3-24 a b^2+16 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}+\frac{(7 a-6 b) \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{24 f}+\frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f} \]

[Out]

((a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f - ((a^3 + 6*a^2*b - 24*a*b^2 +
16*b^3)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(16*b^(3/2)*f) + ((a^2 - 10*a*b + 8*b^2)*
Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(16*b*f) + ((7*a - 6*b)*Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(2
4*f) + (b*Tan[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(6*f)

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Rubi [A]  time = 0.354926, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3670, 477, 582, 523, 217, 206, 377, 203} \[ \frac{\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{16 b f}-\frac{\left (6 a^2 b+a^3-24 a b^2+16 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}+\frac{(7 a-6 b) \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{24 f}+\frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b)^(3/2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/f - ((a^3 + 6*a^2*b - 24*a*b^2 +
16*b^3)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(16*b^(3/2)*f) + ((a^2 - 10*a*b + 8*b^2)*
Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(16*b*f) + ((7*a - 6*b)*Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(2
4*f) + (b*Tan[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(6*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 477

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*(e*x)
^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(b*e*(m + n*(p + q) + 1)), x] + Dist[1/(b*(m + n*(p + q) + 1
)), Int[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*((c*b - a*d)*(m + 1) + c*b*n*(p + q)) + (d*(c*b - a*d
)*(m + 1) + d*n*(q - 1)*(b*c - a*d) + c*b*d*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && N
eQ[b*c - a*d, 0] && IGtQ[n, 0] && GtQ[q, 1] && 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) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}+\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a (6 a-5 b)+(7 a-6 b) b x^2\right )}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{6 f}\\ &=\frac{(7 a-6 b) \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{24 f}+\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a (7 a-6 b) b-3 b \left (a^2-10 a b+8 b^2\right ) x^2\right )}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{24 b f}\\ &=\frac{\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{16 b f}+\frac{(7 a-6 b) \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{24 f}+\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}+\frac{\operatorname{Subst}\left (\int \frac{-3 a b \left (a^2-10 a b+8 b^2\right )-3 b \left (a^3+6 a^2 b-24 a b^2+16 b^3\right ) x^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{48 b^2 f}\\ &=\frac{\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{16 b f}+\frac{(7 a-6 b) \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{24 f}+\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}+\frac{(a-b)^2 \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^3+6 a^2 b-24 a b^2+16 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 b f}\\ &=\frac{\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{16 b f}+\frac{(7 a-6 b) \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{24 f}+\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}+\frac{(a-b)^2 \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^3+6 a^2 b-24 a b^2+16 b^3\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 )}{16 b f}\\ &=\frac{(a-b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{f}-\frac{\left (a^3+6 a^2 b-24 a b^2+16 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac{\left (a^2-10 a b+8 b^2\right ) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{16 b f}+\frac{(7 a-6 b) \tan ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{24 f}+\frac{b \tan ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{6 f}\\ \end{align*}

Mathematica [C]  time = 6.35628, size = 833, normalized size = 3.72 \[ \frac{\sqrt{\frac{\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac{1}{6} b \tan (e+f x) \sec ^4(e+f x)+\frac{7}{24} (a \sin (e+f x)-2 b \sin (e+f x)) \sec ^3(e+f x)+\frac{\left (3 \sin (e+f x) a^2-44 b \sin (e+f x) a+44 b^2 \sin (e+f x)\right ) \sec (e+f x)}{48 b}\right )}{f}-\frac{-\frac{b \left (a^3-2 b a^2-8 b^2 a+8 b^3\right ) \sqrt{\frac{a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right ),1\right ) \sin ^4(e+f x)}{a (a+b+(a-b) \cos (2 (e+f x)))}-\frac{4 b \left (-8 b^3+16 a b^2-8 a^2 b\right ) \sqrt{\cos (2 (e+f x))+1} \sqrt{\frac{a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac{\sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right ),1\right ) \sin ^4(e+f x)}{4 a \sqrt{\cos (2 (e+f x))+1} \sqrt{a+b+(a-b) \cos (2 (e+f x))}}-\frac{\sqrt{-\frac{a \cot ^2(e+f x)}{b}} \sqrt{-\frac{a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \csc (2 (e+f x)) \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right ) \sin ^4(e+f x)}{2 (a-b) \sqrt{\cos (2 (e+f x))+1} \sqrt{a+b+(a-b) \cos (2 (e+f x))}}\right )}{\sqrt{a+b+(a-b) \cos (2 (e+f x))}}}{8 b f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-(-((b*(a^3 - 2*a^2*b - 8*a*b^2 + 8*b^3)*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*(-8*a^2*b + 16*a*b^2 -
8*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)*Co
s[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)]])/(8*b*f) + (Sqrt[(a + b + a*Cos[2*(e +
f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*((7*Sec[e + f*x]^3*(a*Sin[e + f*x] - 2*b*Sin[e + f*x]))/2
4 + (Sec[e + f*x]*(3*a^2*Sin[e + f*x] - 44*a*b*Sin[e + f*x] + 44*b^2*Sin[e + f*x]))/(48*b) + (b*Sec[e + f*x]^4
*Tan[e + f*x])/6))/f

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Maple [B]  time = 0.017, size = 510, normalized size = 2.3 \begin{align*} \text{result too large to display} \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)^(3/2),x)

[Out]

1/6/f*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(5/2)/b-1/24/f*a/b*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2)-1/16/f*a^2/b*(a+b*t
an(f*x+e)^2)^(1/2)*tan(f*x+e)-1/16/f*a^3/b^(3/2)*ln(b^(1/2)*tan(f*x+e)+(a+b*tan(f*x+e)^2)^(1/2))-1/4/f*tan(f*x
+e)*(a+b*tan(f*x+e)^2)^(3/2)-3/8/f*a*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)-3/8/f*a^2/b^(1/2)*ln(b^(1/2)*tan(f*x+
e)+(a+b*tan(f*x+e)^2)^(1/2))+1/2*b*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f+3/2/f*b^(1/2)*a*ln(b^(1/2)*tan(f*x+e)
+(a+b*tan(f*x+e)^2)^(1/2))-1/f*b^(3/2)*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)/(
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))-2/f*a/b*(b^4*(a-b))^(1/2)/(a-b)*a
rctan(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^2*(b^4*(a-b))^(1/2)/b^2/(a-b)*arc
tan(b^2*(a-b)/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{4}\,{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)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^4, x)

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Fricas [A]  time = 21.1036, size = 2103, normalized size = 9.39 \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)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/96*(3*(a^3 + 6*a^2*b - 24*a*b^2 + 16*b^3)*sqrt(b)*log(2*b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqr
t(b)*tan(f*x + e) + a) - 48*(a*b^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*(8*b^3*tan(f*x + e)^5 + 2*(7*a*b^2 - 6*b^3)*ta
n(f*x + e)^3 + 3*(a^2*b - 10*a*b^2 + 8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/48*(3*(a^3 +
6*a^2*b - 24*a*b^2 + 16*b^3)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) - 24*(a*b^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)) + (8*b^3*tan(f*x + e)^5 + 2*(7*a*b^2 - 6*b^3)*tan(f*x + e)^3 + 3*(a^2*b - 10*a*b^2 +
8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/96*(96*(a*b^2 - b^3)*sqrt(a - b)*arctan(-sqrt(b*t
an(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + 3*(a^3 + 6*a^2*b - 24*a*b^2 + 16*b^3)*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*(8*b^3*tan(f*x + e)^5 + 2*(7*a*b^2 - 6*b^
3)*tan(f*x + e)^3 + 3*(a^2*b - 10*a*b^2 + 8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b^2*f), 1/48*(48*(
a*b^2 - b^3)*sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + 3*(a^3 + 6*a^2*b - 2
4*a*b^2 + 16*b^3)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/(b*tan(f*x + e))) + (8*b^3*tan(f*x + e)^
5 + 2*(7*a*b^2 - 6*b^3)*tan(f*x + e)^3 + 3*(a^2*b - 10*a*b^2 + 8*b^3)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a)
)/(b^2*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan ^{4}{\left (e + f x \right )}\, 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)**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x)**2)**(3/2)*tan(e + f*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{4}\,{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)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^4, x)