3.344 \(\int \frac {\tan ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=59 \[ -\frac {(a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a b^{3/2} f}+\frac {x}{a}+\frac {\tan (e+f x)}{b f} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4141, 1975, 479, 522, 203, 205} \[ -\frac {(a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a b^{3/2} f}+\frac {x}{a}+\frac {\tan (e+f x)}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/a - ((a + b)^(3/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*b^(3/2)*f) + Tan[e + f*x]/(b*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 205

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

Rule 479

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 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

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

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Mathematica [C]  time = 1.10, size = 206, normalized size = 3.49 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt {a+b} \sqrt {b (\sin (e)+i \cos (e))^4} (a \sec (e) \sin (f x) \sec (e+f x)+b f x)+(a+b)^2 (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )\right )}{2 a b f \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*((a + b)^2*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b
)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt
[a + b]*Sqrt[b*(I*Cos[e] + Sin[e])^4]*(b*f*x + a*Sec[e]*Sec[e + f*x]*Sin[f*x])))/(2*a*b*Sqrt[a + b]*f*(a + b*S
ec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])

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fricas [B]  time = 0.58, size = 297, normalized size = 5.03 \[ \left [\frac {4 \, b f x \cos \left (f x + e\right ) + {\left (a + b\right )} \sqrt {-\frac {a + b}{b}} \cos \left (f x + e\right ) \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, a \sin \left (f x + e\right )}{4 \, a b f \cos \left (f x + e\right )}, \frac {2 \, b f x \cos \left (f x + e\right ) + {\left (a + b\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) + 2 \, a \sin \left (f x + e\right )}{2 \, a b f \cos \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*b*f*x*cos(f*x + e) + (a + b)*sqrt(-(a + b)/b)*cos(f*x + e)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 -
2*(3*a*b + 4*b^2)*cos(f*x + e)^2 + 4*((a*b + 2*b^2)*cos(f*x + e)^3 - b^2*cos(f*x + e))*sqrt(-(a + b)/b)*sin(f
*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + 4*a*sin(f*x + e))/(a*b*f*cos(f*x + e)), 1/
2*(2*b*f*x*cos(f*x + e) + (a + b)*sqrt((a + b)/b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt((a + b)/b)/((
a + b)*cos(f*x + e)*sin(f*x + e)))*cos(f*x + e) + 2*a*sin(f*x + e))/(a*b*f*cos(f*x + e))]

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giac [A]  time = 3.03, size = 91, normalized size = 1.54 \[ \frac {\frac {f x + e}{a} + \frac {\tan \left (f x + e\right )}{b} - \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a^{2} + 2 \, a b + b^{2}\right )}}{\sqrt {a b + b^{2}} a b}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((f*x + e)/a + tan(f*x + e)/b - (pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))
*(a^2 + 2*a*b + b^2)/(sqrt(a*b + b^2)*a*b))/f

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maple [B]  time = 0.68, size = 121, normalized size = 2.05 \[ \frac {\tan \left (f x +e \right )}{b f}-\frac {a \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f b \sqrt {\left (a +b \right ) b}}-\frac {2 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \sqrt {\left (a +b \right ) b}}-\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f a \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 66, normalized size = 1.12 \[ \frac {\frac {f x + e}{a} + \frac {\tan \left (f x + e\right )}{b} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a b}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.68, size = 410, normalized size = 6.95 \[ \frac {\mathrm {atan}\left (\frac {8\,a^2\,\mathrm {tan}\left (e+f\,x\right )}{12\,a\,b+8\,a^2+6\,b^2+\frac {2\,a^3}{b}}+\frac {2\,a^3\,\mathrm {tan}\left (e+f\,x\right )}{2\,a^3+8\,a^2\,b+12\,a\,b^2+6\,b^3}+\frac {6\,b^2\,\mathrm {tan}\left (e+f\,x\right )}{12\,a\,b+8\,a^2+6\,b^2+\frac {2\,a^3}{b}}+\frac {12\,a\,b\,\mathrm {tan}\left (e+f\,x\right )}{12\,a\,b+8\,a^2+6\,b^2+\frac {2\,a^3}{b}}\right )}{a\,f}+\frac {\mathrm {tan}\left (e+f\,x\right )}{b\,f}+\frac {\mathrm {atanh}\left (\frac {6\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}}{18\,a\,b^2+20\,a^2\,b+10\,a^3+6\,b^3+\frac {2\,a^4}{b}}+\frac {6\,a\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}}{2\,a^4+10\,a^3\,b+20\,a^2\,b^2+18\,a\,b^3+6\,b^4}+\frac {2\,a^2\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^3\,b^3-3\,a^2\,b^4-3\,a\,b^5-b^6}}{2\,a^4\,b+10\,a^3\,b^2+20\,a^2\,b^3+18\,a\,b^4+6\,b^5}\right )\,\sqrt {-b^3\,{\left (a+b\right )}^3}}{a\,b^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((8*a^2*tan(e + f*x))/(12*a*b + 8*a^2 + 6*b^2 + (2*a^3)/b) + (2*a^3*tan(e + f*x))/(12*a*b^2 + 8*a^2*b + 2*
a^3 + 6*b^3) + (6*b^2*tan(e + f*x))/(12*a*b + 8*a^2 + 6*b^2 + (2*a^3)/b) + (12*a*b*tan(e + f*x))/(12*a*b + 8*a
^2 + 6*b^2 + (2*a^3)/b))/(a*f) + tan(e + f*x)/(b*f) + (atanh((6*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*b^4 - a^
3*b^3)^(1/2))/(18*a*b^2 + 20*a^2*b + 10*a^3 + 6*b^3 + (2*a^4)/b) + (6*a*tan(e + f*x)*(- 3*a*b^5 - b^6 - 3*a^2*
b^4 - a^3*b^3)^(1/2))/(18*a*b^3 + 10*a^3*b + 2*a^4 + 6*b^4 + 20*a^2*b^2) + (2*a^2*tan(e + f*x)*(- 3*a*b^5 - b^
6 - 3*a^2*b^4 - a^3*b^3)^(1/2))/(18*a*b^4 + 2*a^4*b + 6*b^5 + 20*a^2*b^3 + 10*a^3*b^2))*(-b^3*(a + b)^3)^(1/2)
)/(a*b^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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