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

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

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 169, 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, 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 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 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 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 ^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*}

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Mathematica [C]  time = 6.25, size = 767, normalized size = 4.54 \[ \frac {\sqrt {\frac {a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b}{\cos (2 (e+f x))+1}} \left (\frac {\sec (e+f x) (a \sin (e+f x)-6 b \sin (e+f x))}{8 b}+\frac {1}{4} \tan (e+f x) \sec ^2(e+f x)\right )}{f}-\frac {-\frac {b \left (a^2-4 b^2\right ) \sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {\frac {(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{a ((a-b) \cos (2 (e+f x))+a+b)}-\frac {4 b \left (4 b^2-4 a b\right ) \sqrt {\cos (2 (e+f x))+1} \sqrt {\frac {(a-b) \cos (2 (e+f x))+a+b}{\cos (2 (e+f x))+1}} \left (\frac {\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{4 a \sqrt {\cos (2 (e+f x))+1} \sqrt {(a-b) \cos (2 (e+f x))+a+b}}-\frac {\sin ^4(e+f x) \csc (2 (e+f x)) \sqrt {-\frac {a \cot ^2(e+f x)}{b}} \sqrt {-\frac {a (\cos (2 (e+f x))+1) \csc ^2(e+f x)}{b}} \sqrt {\frac {\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac {b}{a-b};\left .\sin ^{-1}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{2 (a-b) \sqrt {\cos (2 (e+f x))+1} \sqrt {(a-b) \cos (2 (e+f x))+a+b}}\right )}{\sqrt {(a-b) \cos (2 (e+f x))+a+b}}}{4 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(-((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
)]*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)]])/(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

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fricas [A]  time = 1.20, size = 671, normalized size = 3.97 \[ \left [\frac {8 \, \sqrt {-a + b} b^{2} \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 (a^{2} + 4 \, 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 (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^{2} f}, \frac {16 \, \sqrt {a - b} b^{2} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) - {\left (a^{2} + 4 \, 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 (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^{2} f}, \frac {4 \, \sqrt {-a + b} b^{2} \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 (a^{2} + 4 \, 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 (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^{2} f}, \frac {8 \, \sqrt {a - b} b^{2} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{\sqrt {a - b} \tan \left (f x + e\right )}\right ) + {\left (a^{2} + 4 \, 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 (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^{2} f}\right ] \]

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)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{4}\,{d x} \]

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)

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maple [B]  time = 0.38, size = 323, normalized size = 1.91 \[ \frac {\tan \left (f x +e \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{4 f b}-\frac {a \tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{8 f b}-\frac {a^{2} \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{8 f \,b^{\frac {3}{2}}}-\frac {\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )}{2 f}-\frac {a \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 f \sqrt {b}}+\frac {\sqrt {b}\, \ln \left (\tan \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{f}-\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f b \left (a -b \right )}+\frac {a \sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {\left (a -b \right ) b^{2} \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{f \,b^{2} \left (a -b \right )} \]

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/b*a*tan(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)-1/8/f/b^(3/2)*a^2*ln
(tan(f*x+e)*b^(1/2)+(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(tan
(f*x+e)*b^(1/2)+(a+b*tan(f*x+e)^2)^(1/2))+1/f*b^(1/2)*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)/b/(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*(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 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \tan \left (f x + e\right )^{4}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^4*(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(tan(e + f*x)^4*(a + b*tan(e + f*x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \tan ^{4}{\left (e + f x \right )}\, dx \]

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)

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