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

Optimal. Leaf size=90 \[ \frac {(a-2 b) \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 b^{3/2} f}+\frac {x}{a^2}-\frac {(a+b) \tan (e+f x)}{2 a b f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 90, 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, 470, 522, 203, 205} \[ \frac {(a-2 b) \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 b^{3/2} f}+\frac {x}{a^2}-\frac {(a+b) \tan (e+f x)}{2 a b f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 470

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

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Mathematica [C]  time = 2.56, size = 249, normalized size = 2.77 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac {\left (-a^2+a b+2 b^2\right ) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{b f \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+2 x (a \cos (2 (e+f x))+a+2 b)+\frac {(a+b) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b f (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}\right )}{8 a^2 \left (a+b \sec ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*(2*x*(a + 2*b + a*Cos[2*(e + f*x)]) + ((-a^2 + a*b + 2*b^2)*Arc
Tan[(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])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[2*e] - I*Sin[2*e]))/(b*Sqrt[a + b]*f*Sqrt[b*(Cos[e] - I
*Sin[e])^4]) + ((a + b)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(8*a^
2*(a + b*Sec[e + f*x]^2)^2)

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fricas [B]  time = 0.53, size = 393, normalized size = 4.37 \[ \left [\frac {8 \, a b f x \cos \left (f x + e\right )^{2} + 8 \, b^{2} f x - 4 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt {-\frac {a + b}{b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}, \frac {4 \, a b f x \cos \left (f x + e\right )^{2} + 4 \, b^{2} f x - 2 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.76, size = 126, normalized size = 1.40 \[ \frac {\frac {2 \, {\left (f x + e\right )}}{a^{2}} + \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a^{2} - a b - 2 \, b^{2}\right )}}{\sqrt {a b + b^{2}} a^{2} b} - \frac {a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a b}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.90, size = 168, normalized size = 1.87 \[ -\frac {\tan \left (f x +e \right )}{2 f b \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f b \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f a \sqrt {\left (a +b \right ) b}}-\frac {\tan \left (f x +e \right )}{2 a f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{2} \sqrt {\left (a +b \right ) b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 96, normalized size = 1.07 \[ -\frac {\frac {{\left (a + b\right )} \tan \left (f x + e\right )}{a b^{2} \tan \left (f x + e\right )^{2} + a^{2} b + a b^{2}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}} - \frac {{\left (a^{2} - a b - 2 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2} b}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.78, size = 285, normalized size = 3.17 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (e+f\,x\right )}{\frac {3\,b}{2\,a}-\frac {a}{2\,b}+1}-\frac {\mathrm {tan}\left (e+f\,x\right )}{2\,\left (\frac {b}{a}+\frac {3\,b^2}{2\,a^2}-\frac {1}{2}\right )}+\frac {3\,b\,\mathrm {tan}\left (e+f\,x\right )}{2\,\left (a+\frac {3\,b}{2}-\frac {a^2}{2\,b}\right )}\right )}{a^2\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+b\right )}{2\,a\,b\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {\mathrm {atanh}\left (\frac {3\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{2\,\left (\frac {a\,b}{4}-a^2+\frac {3\,b^2}{2}+\frac {a^3}{4\,b}\right )}-\frac {5\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{4\,\left (\frac {a^2}{4}-a\,b+\frac {b^2}{4}+\frac {3\,b^3}{2\,a}\right )}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^4-a\,b^3}}{4\,\left (\frac {a\,b}{4}-b^2+\frac {b^3}{4\,a}+\frac {3\,b^4}{2\,a^2}\right )}\right )\,\sqrt {-b^3\,\left (a+b\right )}\,\left (a-2\,b\right )}{2\,a^2\,b^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(tan(e + f*x)/((3*b)/(2*a) - a/(2*b) + 1) - tan(e + f*x)/(2*(b/a + (3*b^2)/(2*a^2) - 1/2)) + (3*b*tan(e +
f*x))/(2*(a + (3*b)/2 - a^2/(2*b))))/(a^2*f) - (tan(e + f*x)*(a + b))/(2*a*b*f*(a + b + b*tan(e + f*x)^2)) - (
atanh((3*tan(e + f*x)*(- a*b^3 - b^4)^(1/2))/(2*((a*b)/4 - a^2 + (3*b^2)/2 + a^3/(4*b))) - (5*tan(e + f*x)*(-
a*b^3 - b^4)^(1/2))/(4*(a^2/4 - a*b + b^2/4 + (3*b^3)/(2*a))) + (tan(e + f*x)*(- a*b^3 - b^4)^(1/2))/(4*((a*b)
/4 - b^2 + b^3/(4*a) + (3*b^4)/(2*a^2))))*(-b^3*(a + b))^(1/2)*(a - 2*b))/(2*a^2*b^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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