3.343 \(\int \frac {\tan ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=83 \[ \frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a b^{5/2} f}-\frac {(a+2 b) \tan (e+f x)}{b^2 f}-\frac {x}{a}+\frac {\tan ^3(e+f x)}{3 b f} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4141, 1975, 479, 582, 522, 203, 205} \[ \frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a b^{5/2} f}-\frac {(a+2 b) \tan (e+f x)}{b^2 f}-\frac {x}{a}+\frac {\tan ^3(e+f x)}{3 b f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]

[Out]

-(x/a) + ((a + b)^(5/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*b^(5/2)*f) - ((a + 2*b)*Tan[e + f*x])/(
b^2*f) + Tan[e + f*x]^3/(3*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 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 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 ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{\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^6}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan ^3(e+f x)}{3 b f}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 (a+b)+3 (a+2 b) x^2\right )}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{3 b f}\\ &=-\frac {(a+2 b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f}+\frac {\operatorname {Subst}\left (\int \frac {3 (a+b) (a+2 b)+3 \left (a^2+3 a b+3 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{3 b^2 f}\\ &=-\frac {(a+2 b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a f}+\frac {(a+b)^3 \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a b^2 f}\\ &=-\frac {x}{a}+\frac {(a+b)^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a b^{5/2} f}-\frac {(a+2 b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f}\\ \end {align*}

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Mathematica [C]  time = 2.79, size = 229, normalized size = 2.76 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-\frac {(3 a+7 b) \sec (e) \sin (f x) \sec (e+f x)}{b^2 f}-\frac {3 (a+b)^{5/2} (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{a b^2 f \sqrt {b (\cos (e)-i \sin (e))^4}}-\frac {3 x}{a}+\frac {\sec (e) \sin (f x) \sec ^3(e+f x)}{b f}+\frac {\tan (e) \sec ^2(e+f x)}{b f}\right )}{6 \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*((-3*x)/a - (3*(a + b)^(5/2)*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]))/(a*b^2*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) - ((3*a + 7*b)*Sec[e]*Sec[e + f*x]*Sin[f*x])/(b^2*f) + (S
ec[e]*Sec[e + f*x]^3*Sin[f*x])/(b*f) + (Sec[e + f*x]^2*Tan[e])/(b*f)))/(6*(a + b*Sec[e + f*x]^2))

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fricas [B]  time = 0.54, size = 373, normalized size = 4.49 \[ \left [-\frac {12 \, b^{2} f x \cos \left (f x + e\right )^{3} - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {a + b}{b}} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 4 \, {\left ({\left (3 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sin \left (f x + e\right )}{12 \, a b^{2} f \cos \left (f x + e\right )^{3}}, -\frac {6 \, b^{2} f x \cos \left (f x + e\right )^{3} + 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} + 2 \, {\left ({\left (3 \, a^{2} + 7 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sin \left (f x + e\right )}{6 \, a b^{2} f \cos \left (f x + e\right )^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/12*(12*b^2*f*x*cos(f*x + e)^3 - 3*(a^2 + 2*a*b + b^2)*sqrt(-(a + b)/b)*cos(f*x + e)^3*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*((3*a^2 + 7*a*b)*
cos(f*x + e)^2 - a*b)*sin(f*x + e))/(a*b^2*f*cos(f*x + e)^3), -1/6*(6*b^2*f*x*cos(f*x + e)^3 + 3*(a^2 + 2*a*b
+ 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)))*cos(f*x + e)^3 + 2*((3*a^2 + 7*a*b)*cos(f*x + e)^2 - a*b)*sin(f*x + e))/(a*b^2*f*cos(f*x + e)^3)]

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giac [A]  time = 6.31, size = 132, normalized size = 1.59 \[ -\frac {\frac {3 \, {\left (f x + e\right )}}{a} - \frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a b^{2}} - \frac {b^{2} \tan \left (f x + e\right )^{3} - 3 \, a b \tan \left (f x + e\right ) - 6 \, b^{2} \tan \left (f x + e\right )}{b^{3}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-1/3*(3*(f*x + e)/a - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*
x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a*b^2) - (b^2*tan(f*x + e)^3 - 3*a*b*tan(f*x + e) - 6*b^2*tan(f*x +
e))/b^3)/f

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maple [B]  time = 0.84, size = 186, normalized size = 2.24 \[ \frac {\tan ^{3}\left (f x +e \right )}{3 b f}-\frac {a \tan \left (f x +e \right )}{f \,b^{2}}-\frac {2 \tan \left (f x +e \right )}{b f}+\frac {a^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,b^{2} \sqrt {\left (a +b \right ) b}}+\frac {3 a \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f b \sqrt {\left (a +b \right ) b}}+\frac {3 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \sqrt {\left (a +b \right ) b}}+\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f a \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^6/(a+b*sec(f*x+e)^2),x)

[Out]

1/3*tan(f*x+e)^3/b/f-1/f/b^2*a*tan(f*x+e)-2*tan(f*x+e)/b/f+1/f/b^2*a^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a
+b)*b)^(1/2))+3/f/b*a/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+3/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 = 95, normalized size = 1.14 \[ -\frac {\frac {3 \, {\left (f x + e\right )}}{a} - \frac {b \tan \left (f x + e\right )^{3} - 3 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )}{b^{2}} - \frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a b^{2}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.92, size = 1109, normalized size = 13.36 \[ \frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,b\,f}-\frac {\mathrm {atan}\left (\frac {40\,a^2\,\mathrm {tan}\left (e+f\,x\right )}{30\,a\,b+40\,a^2+10\,b^2+\frac {30\,a^3}{b}+\frac {12\,a^4}{b^2}+\frac {2\,a^5}{b^3}}+\frac {30\,a^3\,\mathrm {tan}\left (e+f\,x\right )}{30\,a\,b^2+40\,a^2\,b+30\,a^3+10\,b^3+\frac {12\,a^4}{b}+\frac {2\,a^5}{b^2}}+\frac {12\,a^4\,\mathrm {tan}\left (e+f\,x\right )}{30\,a\,b^3+30\,a^3\,b+12\,a^4+10\,b^4+40\,a^2\,b^2+\frac {2\,a^5}{b}}+\frac {2\,a^5\,\mathrm {tan}\left (e+f\,x\right )}{2\,a^5+12\,a^4\,b+30\,a^3\,b^2+40\,a^2\,b^3+30\,a\,b^4+10\,b^5}+\frac {10\,b^2\,\mathrm {tan}\left (e+f\,x\right )}{30\,a\,b+40\,a^2+10\,b^2+\frac {30\,a^3}{b}+\frac {12\,a^4}{b^2}+\frac {2\,a^5}{b^3}}+\frac {30\,a\,b\,\mathrm {tan}\left (e+f\,x\right )}{30\,a\,b+40\,a^2+10\,b^2+\frac {30\,a^3}{b}+\frac {12\,a^4}{b^2}+\frac {2\,a^5}{b^3}}\right )}{a\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+2\,b\right )}{b^2\,f}-\frac {\mathrm {atan}\left (\frac {\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {2\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+2\,b^6\right )}{b^3}+\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {4\,a^4\,b^3+12\,a^3\,b^4+8\,a^2\,b^5}{b^3}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (4\,a^3\,b^5+8\,a^2\,b^6\right )\,\sqrt {-b^5\,{\left (a+b\right )}^5}}{a\,b^8}\right )}{2\,a\,b^5}\right )\,1{}\mathrm {i}}{2\,a\,b^5}+\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {2\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+2\,b^6\right )}{b^3}-\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {4\,a^4\,b^3+12\,a^3\,b^4+8\,a^2\,b^5}{b^3}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (4\,a^3\,b^5+8\,a^2\,b^6\right )\,\sqrt {-b^5\,{\left (a+b\right )}^5}}{a\,b^8}\right )}{2\,a\,b^5}\right )\,1{}\mathrm {i}}{2\,a\,b^5}}{\frac {2\,\left (a^5+6\,a^4\,b+15\,a^3\,b^2+19\,a^2\,b^3+12\,a\,b^4+3\,b^5\right )}{b^3}-\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {2\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+2\,b^6\right )}{b^3}+\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {4\,a^4\,b^3+12\,a^3\,b^4+8\,a^2\,b^5}{b^3}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (4\,a^3\,b^5+8\,a^2\,b^6\right )\,\sqrt {-b^5\,{\left (a+b\right )}^5}}{a\,b^8}\right )}{2\,a\,b^5}\right )}{2\,a\,b^5}+\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {2\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^6+6\,a^5\,b+15\,a^4\,b^2+20\,a^3\,b^3+15\,a^2\,b^4+6\,a\,b^5+2\,b^6\right )}{b^3}-\frac {\sqrt {-b^5\,{\left (a+b\right )}^5}\,\left (\frac {4\,a^4\,b^3+12\,a^3\,b^4+8\,a^2\,b^5}{b^3}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (4\,a^3\,b^5+8\,a^2\,b^6\right )\,\sqrt {-b^5\,{\left (a+b\right )}^5}}{a\,b^8}\right )}{2\,a\,b^5}\right )}{2\,a\,b^5}}\right )\,\sqrt {-b^5\,{\left (a+b\right )}^5}\,1{}\mathrm {i}}{a\,b^5\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2),x)

[Out]

tan(e + f*x)^3/(3*b*f) - atan((40*a^2*tan(e + f*x))/(30*a*b + 40*a^2 + 10*b^2 + (30*a^3)/b + (12*a^4)/b^2 + (2
*a^5)/b^3) + (30*a^3*tan(e + f*x))/(30*a*b^2 + 40*a^2*b + 30*a^3 + 10*b^3 + (12*a^4)/b + (2*a^5)/b^2) + (12*a^
4*tan(e + f*x))/(30*a*b^3 + 30*a^3*b + 12*a^4 + 10*b^4 + 40*a^2*b^2 + (2*a^5)/b) + (2*a^5*tan(e + f*x))/(30*a*
b^4 + 12*a^4*b + 2*a^5 + 10*b^5 + 40*a^2*b^3 + 30*a^3*b^2) + (10*b^2*tan(e + f*x))/(30*a*b + 40*a^2 + 10*b^2 +
(30*a^3)/b + (12*a^4)/b^2 + (2*a^5)/b^3) + (30*a*b*tan(e + f*x))/(30*a*b + 40*a^2 + 10*b^2 + (30*a^3)/b + (12
*a^4)/b^2 + (2*a^5)/b^3))/(a*f) - (tan(e + f*x)*(a + 2*b))/(b^2*f) - (atan((((-b^5*(a + b)^5)^(1/2)*((2*tan(e
+ f*x)*(6*a*b^5 + 6*a^5*b + a^6 + 2*b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2))/b^3 + ((-b^5*(a + b)^5)^(1/2)
*((8*a^2*b^5 + 12*a^3*b^4 + 4*a^4*b^3)/b^3 + (tan(e + f*x)*(8*a^2*b^6 + 4*a^3*b^5)*(-b^5*(a + b)^5)^(1/2))/(a*
b^8)))/(2*a*b^5))*1i)/(2*a*b^5) + ((-b^5*(a + b)^5)^(1/2)*((2*tan(e + f*x)*(6*a*b^5 + 6*a^5*b + a^6 + 2*b^6 +
15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2))/b^3 - ((-b^5*(a + b)^5)^(1/2)*((8*a^2*b^5 + 12*a^3*b^4 + 4*a^4*b^3)/b^3
- (tan(e + f*x)*(8*a^2*b^6 + 4*a^3*b^5)*(-b^5*(a + b)^5)^(1/2))/(a*b^8)))/(2*a*b^5))*1i)/(2*a*b^5))/((2*(12*a
*b^4 + 6*a^4*b + a^5 + 3*b^5 + 19*a^2*b^3 + 15*a^3*b^2))/b^3 - ((-b^5*(a + b)^5)^(1/2)*((2*tan(e + f*x)*(6*a*b
^5 + 6*a^5*b + a^6 + 2*b^6 + 15*a^2*b^4 + 20*a^3*b^3 + 15*a^4*b^2))/b^3 + ((-b^5*(a + b)^5)^(1/2)*((8*a^2*b^5
+ 12*a^3*b^4 + 4*a^4*b^3)/b^3 + (tan(e + f*x)*(8*a^2*b^6 + 4*a^3*b^5)*(-b^5*(a + b)^5)^(1/2))/(a*b^8)))/(2*a*b
^5)))/(2*a*b^5) + ((-b^5*(a + b)^5)^(1/2)*((2*tan(e + f*x)*(6*a*b^5 + 6*a^5*b + a^6 + 2*b^6 + 15*a^2*b^4 + 20*
a^3*b^3 + 15*a^4*b^2))/b^3 - ((-b^5*(a + b)^5)^(1/2)*((8*a^2*b^5 + 12*a^3*b^4 + 4*a^4*b^3)/b^3 - (tan(e + f*x)
*(8*a^2*b^6 + 4*a^3*b^5)*(-b^5*(a + b)^5)^(1/2))/(a*b^8)))/(2*a*b^5)))/(2*a*b^5)))*(-b^5*(a + b)^5)^(1/2)*1i)/
(a*b^5*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{6}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**6/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(tan(e + f*x)**6/(a + b*sec(e + f*x)**2), x)

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