\(\int (a+b \sec ^2(e+f x)) \tan ^5(e+f x) \, dx\) [311]

Optimal result

Integrand size = 21, antiderivative size = 72 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {a \log (\cos (e+f x))}{f}-\frac {(2 a-b) \sec ^2(e+f x)}{2 f}+\frac {(a-2 b) \sec ^4(e+f x)}{4 f}+\frac {b \sec ^6(e+f x)}{6 f} \]

[Out]

-a*ln(cos(f*x+e))/f-1/2*(2*a-b)*sec(f*x+e)^2/f+1/4*(a-2*b)*sec(f*x+e)^4/f+1/6*b*sec(f*x+e)^6/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4223, 457, 77} \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=\frac {(a-2 b) \sec ^4(e+f x)}{4 f}-\frac {(2 a-b) \sec ^2(e+f x)}{2 f}-\frac {a \log (\cos (e+f x))}{f}+\frac {b \sec ^6(e+f x)}{6 f} \]

[In]

Int[(a + b*Sec[e + f*x]^2)*Tan[e + f*x]^5,x]

[Out]

-((a*Log[Cos[e + f*x]])/f) - ((2*a - b)*Sec[e + f*x]^2)/(2*f) + ((a - 2*b)*Sec[e + f*x]^4)/(4*f) + (b*Sec[e +
f*x]^6)/(6*f)

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 457

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

Rule 4223

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^2 \left (b+a x^2\right )}{x^7} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \frac {(1-x)^2 (b+a x)}{x^4} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {b}{x^4}+\frac {a-2 b}{x^3}+\frac {-2 a+b}{x^2}+\frac {a}{x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {a \log (\cos (e+f x))}{f}-\frac {(2 a-b) \sec ^2(e+f x)}{2 f}+\frac {(a-2 b) \sec ^4(e+f x)}{4 f}+\frac {b \sec ^6(e+f x)}{6 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.76 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=\frac {b \tan ^6(e+f x)}{6 f}-\frac {a \left (4 \log (\cos (e+f x))+2 \tan ^2(e+f x)-\tan ^4(e+f x)\right )}{4 f} \]

[In]

Integrate[(a + b*Sec[e + f*x]^2)*Tan[e + f*x]^5,x]

[Out]

(b*Tan[e + f*x]^6)/(6*f) - (a*(4*Log[Cos[e + f*x]] + 2*Tan[e + f*x]^2 - Tan[e + f*x]^4))/(4*f)

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.76

method	result	size
parts	\(\frac {a \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {b \tan \left (f x +e \right )^{6}}{6 f}\)	\(55\)
derivativedivides	\(\frac {\frac {\sec \left (f x +e \right )^{6} b}{6}+\frac {\sec \left (f x +e \right )^{4} a}{4}-\frac {\sec \left (f x +e \right )^{4} b}{2}-\sec \left (f x +e \right )^{2} a +\frac {b \sec \left (f x +e \right )^{2}}{2}+a \ln \left (\sec \left (f x +e \right )\right )}{f}\)	\(70\)
default	\(\frac {\frac {\sec \left (f x +e \right )^{6} b}{6}+\frac {\sec \left (f x +e \right )^{4} a}{4}-\frac {\sec \left (f x +e \right )^{4} b}{2}-\sec \left (f x +e \right )^{2} a +\frac {b \sec \left (f x +e \right )^{2}}{2}+a \ln \left (\sec \left (f x +e \right )\right )}{f}\)	\(70\)
risch	\(i a x +\frac {2 i a e}{f}+\frac {-4 a \,{\mathrm e}^{10 i \left (f x +e \right )}+2 b \,{\mathrm e}^{10 i \left (f x +e \right )}-12 a \,{\mathrm e}^{8 i \left (f x +e \right )}-16 a \,{\mathrm e}^{6 i \left (f x +e \right )}+\frac {20 b \,{\mathrm e}^{6 i \left (f x +e \right )}}{3}-12 a \,{\mathrm e}^{4 i \left (f x +e \right )}-4 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{6}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}\)	\(148\)

[In]

int((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x,method=_RETURNVERBOSE)

[Out]

a/f*(1/4*tan(f*x+e)^4-1/2*tan(f*x+e)^2+1/2*ln(1+tan(f*x+e)^2))+1/6*b/f*tan(f*x+e)^6

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.96 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {12 \, a \cos \left (f x + e\right )^{6} \log \left (-\cos \left (f x + e\right )\right ) + 6 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{4} - 3 \, {\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, b}{12 \, f \cos \left (f x + e\right )^{6}} \]

[In]

integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

-1/12*(12*a*cos(f*x + e)^6*log(-cos(f*x + e)) + 6*(2*a - b)*cos(f*x + e)^4 - 3*(a - 2*b)*cos(f*x + e)^2 - 2*b)
/(f*cos(f*x + e)^6)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (56) = 112\).

Time = 0.55 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.61 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=\begin {cases} \frac {a \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {a \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac {b \tan ^{4}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{6 f} - \frac {b \tan ^{2}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{6 f} + \frac {b \sec ^{2}{\left (e + f x \right )}}{6 f} & \text {for}\: f \neq 0 \\x \left (a + b \sec ^{2}{\left (e \right )}\right ) \tan ^{5}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

integrate((a+b*sec(f*x+e)**2)*tan(f*x+e)**5,x)

[Out]

Piecewise((a*log(tan(e + f*x)**2 + 1)/(2*f) + a*tan(e + f*x)**4/(4*f) - a*tan(e + f*x)**2/(2*f) + b*tan(e + f*
x)**4*sec(e + f*x)**2/(6*f) - b*tan(e + f*x)**2*sec(e + f*x)**2/(6*f) + b*sec(e + f*x)**2/(6*f), Ne(f, 0)), (x
*(a + b*sec(e)**2)*tan(e)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.32 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=-\frac {6 \, a \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac {6 \, {\left (2 \, a - b\right )} \sin \left (f x + e\right )^{4} - 3 \, {\left (7 \, a - 2 \, b\right )} \sin \left (f x + e\right )^{2} + 9 \, a - 2 \, b}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{12 \, f} \]

[In]

integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

-1/12*(6*a*log(sin(f*x + e)^2 - 1) - (6*(2*a - b)*sin(f*x + e)^4 - 3*(7*a - 2*b)*sin(f*x + e)^2 + 9*a - 2*b)/(
sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(f*x + e)^2 - 1))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (66) = 132\).

Time = 1.59 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.89 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=\frac {6 \, a \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right ) - 6 \, a \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right ) + \frac {11 \, a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{3} + 90 \, a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{2} + 276 \, a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 280 \, a - 128 \, b}{{\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}^{3}}}{12 \, f} \]

[In]

integrate((a+b*sec(f*x+e)^2)*tan(f*x+e)^5,x, algorithm="giac")

[Out]

1/12*(6*a*log(abs(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2)) - 6*a*l
og(abs(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2)) + (11*a*((cos(f*x
+ e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^3 + 90*a*((cos(f*x + e) + 1)/(cos(f*x +
e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^2 + 276*a*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x +
e) - 1)/(cos(f*x + e) + 1)) + 280*a - 128*b)/((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos
(f*x + e) + 1) + 2)^3)/f

Mupad [B] (verification not implemented)

Time = 19.69 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.72 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^5(e+f x) \, dx=\frac {\frac {a\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2}-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^4}{4}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^6}{6}}{f} \]

[In]

int(tan(e + f*x)^5*(a + b/cos(e + f*x)^2),x)

[Out]

((a*log(tan(e + f*x)^2 + 1))/2 - (a*tan(e + f*x)^2)/2 + (a*tan(e + f*x)^4)/4 + (b*tan(e + f*x)^6)/6)/f