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

Optimal. Leaf size=172 \[ \frac {\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {(5 a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f}+\frac {b \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f} \]

[Out]

-(a-b)^(3/2)*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f+1/8*(3*a^2-12*a*b+8*b^2)*arctanh(b^(1/2
)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/f/b^(1/2)+1/8*(5*a-4*b)*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/f+1/4*b*(a+
b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)^3/f

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Rubi [A]  time = 0.25, antiderivative size = 172, 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, 477, 582, 523, 217, 206, 377, 203} \[ \frac {\left (3 a^2-12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {b \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {(5 a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}-\frac {(a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2*(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) + ((3*a^2 - 12*a*b + 8*b^2)
*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*Sqrt[b]*f) + ((5*a - 4*b)*Tan[e + f*x]*Sqrt[a
+ b*Tan[e + f*x]^2])/(8*f) + (b*Tan[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2])/(4*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 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 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 \tan ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {b \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a (4 a-3 b)+(5 a-4 b) b x^2\right )}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {(5 a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {b \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}-\frac {\operatorname {Subst}\left (\int \frac {a (5 a-4 b) b-b \left (3 a^2-12 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 {(5 a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {b \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 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 (3 a^2-12 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 f}\\ &=\frac {(5 a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {b \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 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 (3 a^2-12 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 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 (3 a^2-12 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 \sqrt {b} f}+\frac {(5 a-4 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{8 f}+\frac {b \tan ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{4 f}\\ \end {align*}

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Mathematica [C]  time = 6.31, size = 771, normalized size = 4.48 \[ \frac {\frac {b \left (a^2+4 a b-4 b^2\right ) \sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {\frac {(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{a ((a-b) \cos (2 (e+f x))+a+b)}+\frac {4 b \left (4 a^2-8 a b+4 b^2\right ) \sqrt {\cos (2 (e+f x))+1} \sqrt {\frac {(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \left (\frac {\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{4 a \sqrt {\cos (2 (e+f x))+1} \sqrt {(a-b) \cos (2 (e+f x))+a+b}}-\frac {\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{2 (a-b) \sqrt {\cos (2 (e+f x))+1} \sqrt {(a-b) \cos (2 (e+f x))+a+b}}\right )}{\sqrt {(a-b) \cos (2 (e+f x))+a+b}}}{4 f}+\frac {\sqrt {\frac {a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b}{\cos (2 (e+f x))+1}} \left (\frac {1}{8} \sec (e+f x) (5 a \sin (e+f x)-6 b \sin (e+f x))+\frac {1}{4} b \tan (e+f x) \sec ^2(e+f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*(a^2 + 4*a*b - 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]/Sqr
t[2]], 1]*Sin[e + f*x]^4)/(a*(a + b + (a - b)*Cos[2*(e + f*x)])) + (4*b*(4*a^2 - 8*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)]*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)]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1]*S
in[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)]])/(4*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]*(5*a*Sin[e + f*x] - 6*b*Sin[e + f*x]))/8 + (b*Sec[e + f*x]^2*Tan[e
+ f*x])/4))/f

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fricas [A]  time = 1.24, size = 708, normalized size = 4.12 \[ \left [\frac {{\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) - 8 \, {\left (a b - b^{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 (2 \, b^{2} \tan \left (f x + e\right )^{3} + {\left (5 \, a b - 4 \, b^{2}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{16 \, b f}, -\frac {{\left (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) + 4 \, {\left (a b - b^{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 ) - {\left (2 \, b^{2} \tan \left (f x + e\right )^{3} + {\left (5 \, a b - 4 \, b^{2}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{8 \, b f}, -\frac {16 \, {\left (a b - b^{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 (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \sqrt {b} \log \left (2 \, b \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {b} \tan \left (f x + e\right ) + a\right ) - 2 \, {\left (2 \, b^{2} \tan \left (f x + e\right )^{3} + {\left (5 \, a b - 4 \, b^{2}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{16 \, b f}, -\frac {8 \, {\left (a b - b^{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 (3 \, a^{2} - 12 \, a b + 8 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-b}}{b \tan \left (f x + e\right )}\right ) - {\left (2 \, b^{2} \tan \left (f x + e\right )^{3} + {\left (5 \, a b - 4 \, b^{2}\right )} \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a}}{8 \, b f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((3*a^2 - 12*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) - 8*(a*b - b^2)*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*(2*b^2*tan(f*x + e)^3 + (5*a*b - 4*b^2)*tan(f*x + e))*sqrt(b*t
an(f*x + e)^2 + a))/(b*f), -1/8*((3*a^2 - 12*a*b + 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-b)/
(b*tan(f*x + e))) + 4*(a*b - b^2)*sqrt(-a + b)*log(-((a - 2*b)*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*s
qrt(-a + b)*tan(f*x + e) - a)/(tan(f*x + e)^2 + 1)) - (2*b^2*tan(f*x + e)^3 + (5*a*b - 4*b^2)*tan(f*x + e))*sq
rt(b*tan(f*x + e)^2 + a))/(b*f), -1/16*(16*(a*b - b^2)*sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a
- b)*tan(f*x + e))) - (3*a^2 - 12*a*b + 8*b^2)*sqrt(b)*log(2*b*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*s
qrt(b)*tan(f*x + e) + a) - 2*(2*b^2*tan(f*x + e)^3 + (5*a*b - 4*b^2)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))
/(b*f), -1/8*(8*(a*b - b^2)*sqrt(a - b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)/(sqrt(a - b)*tan(f*x + e))) + (3*a^
2 - 12*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 + (5*a*b - 4*b^2)*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a))/(b*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2*(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)^2, x)

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maple [B]  time = 0.31, size = 386, normalized size = 2.24 \[ \frac {\tan \left (f x +e \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{4 f}+\frac {3 a \tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{8 f}+\frac {3 a^{2} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{8 f \sqrt {b}}-\frac {b \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )}{2 f}-\frac {3 \sqrt {b}\, a \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 f}+\frac {b^{\frac {3}{2}} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{f}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \left (a -b \right )}+\frac {2 a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f b \left (a -b \right )}-\frac {a^{2} \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \,b^{2} \left (a -b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2)+3/8/f*a*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)+3/8/f*a^2/b^(1/2)*ln(tan
(f*x+e)*b^(1/2)+(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(tan(f
*x+e)*b^(1/2)+(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*(b^4*(
a-b))^(1/2)/(a-b)*arctan((a-b)*b^2/(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)*arctan((a-b)*b^2/(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)*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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2*(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)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(tan(e + f*x)^2*(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 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2*(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)**2, x)

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