3.4.27 \(\int \frac {\tan ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [327]

Optimal. Leaf size=125 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f} \]

[Out]

-1/2*(a+2*b)*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))/f/(a-b)^(1/2)+1/2*(a+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/b/f

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Rubi [A]
time = 0.09, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3751, 490, 537, 223, 212, 385, 209} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f \sqrt {a-b}}-\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 385

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 490

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(2*n -
1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q) + 1))), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 537

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 3751

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 ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f}-\frac {\text {Subst}\left (\int \frac {a+(a+2 b) x^2}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 b f}\\ &=\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f}+\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}-\frac {(a+2 b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 b f}\\ &=\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f}+\frac {\text {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 {(a+2 b) \text {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 f}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac {\tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 b f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 4.84, size = 271, normalized size = 2.17 \begin {gather*} -\frac {\left (\sqrt {2} a (-a+b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} F\left (\left .\text {ArcSin}\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 \sqrt {2} a b \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \Pi \left (-\frac {b}{a-b};\left .\text {ArcSin}\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-b) (a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)\right ) \tan (e+f x)}{2 \sqrt {2} b (-a+b) f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((Sqrt[2]*a*(-a + b)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcSin[Sqrt[((
a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - 2*Sqrt[2]*a*b*Sqrt[((a + b + (a - b)*Cos[2
*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e
+ f*x]^2)/b]/Sqrt[2]], 1] + (a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2)*Tan[e + f*x])/(Sqrt[2]
*b*(-a + b)*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])

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Maple [A]
time = 0.07, size = 157, normalized size = 1.26

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 b^{\frac {3}{2}}}-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{\sqrt {b}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{b^{2} \left (a -b \right )}}{f}\)	\(157\)
default	\(\frac {\frac {\tan \left (f x +e \right ) \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}{2 b}-\frac {a \ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{2 b^{\frac {3}{2}}}-\frac {\ln \left (\sqrt {b}\, \tan \left (f x +e \right )+\sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}\right )}{\sqrt {b}}+\frac {\sqrt {b^{4} \left (a -b \right )}\, \arctan \left (\frac {b^{2} \left (a -b \right ) \tan \left (f x +e \right )}{\sqrt {b^{4} \left (a -b \right )}\, \sqrt {a +b \left (\tan ^{2}\left (f x +e \right )\right )}}\right )}{b^{2} \left (a -b \right )}}{f}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 5.57, size = 677, normalized size = 5.42 \begin {gather*} \left [-\frac {2 \, \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} + a b - 2 \, 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 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{4 \, {\left (a b^{2} - b^{3}\right )} f}, -\frac {\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} + a b - 2 \, 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 ) - \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{2 \, {\left (a b^{2} - b^{3}\right )} f}, \frac {4 \, \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} + a b - 2 \, 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 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{4 \, {\left (a b^{2} - b^{3}\right )} f}, \frac {2 \, \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} + a b - 2 \, 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 ) + \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{2 \, {\left (a b^{2} - b^{3}\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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

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

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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