\(\int (a+b \sec ^2(e+f x))^2 \tan ^5(e+f x) \, dx\) [324]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=-\frac {a^2 \log (\cos (e+f x))}{f}-\frac {a (a-b) \sec ^2(e+f x)}{f}+\frac {\left (a^2-4 a b+b^2\right ) \sec ^4(e+f x)}{4 f}+\frac {(a-b) b \sec ^6(e+f x)}{3 f}+\frac {b^2 \sec ^8(e+f x)}{8 f} \]

[Out]

-a^2*ln(cos(f*x+e))/f-a*(a-b)*sec(f*x+e)^2/f+1/4*(a^2-4*a*b+b^2)*sec(f*x+e)^4/f+1/3*(a-b)*b*sec(f*x+e)^6/f+1/8
*b^2*sec(f*x+e)^8/f

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4223, 457, 90} \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=\frac {\left (a^2-4 a b+b^2\right ) \sec ^4(e+f x)}{4 f}-\frac {a^2 \log (\cos (e+f x))}{f}+\frac {b (a-b) \sec ^6(e+f x)}{3 f}-\frac {a (a-b) \sec ^2(e+f x)}{f}+\frac {b^2 \sec ^8(e+f x)}{8 f} \]

[In]

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

[Out]

-((a^2*Log[Cos[e + f*x]])/f) - (a*(a - b)*Sec[e + f*x]^2)/f + ((a^2 - 4*a*b + b^2)*Sec[e + f*x]^4)/(4*f) + ((a
- b)*b*Sec[e + f*x]^6)/(3*f) + (b^2*Sec[e + f*x]^8)/(8*f)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -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 )^2}{x^9} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \frac {(1-x)^2 (b+a x)^2}{x^5} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {b^2}{x^5}+\frac {2 (a-b) b}{x^4}+\frac {a^2-4 a b+b^2}{x^3}-\frac {2 a (a-b)}{x^2}+\frac {a^2}{x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {a^2 \log (\cos (e+f x))}{f}-\frac {a (a-b) \sec ^2(e+f x)}{f}+\frac {\left (a^2-4 a b+b^2\right ) \sec ^4(e+f x)}{4 f}+\frac {(a-b) b \sec ^6(e+f x)}{3 f}+\frac {b^2 \sec ^8(e+f x)}{8 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.26 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=-\frac {\cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \left (24 a^2 \log (\cos (e+f x))+24 a (a-b) \sec ^2(e+f x)-6 \left (a^2-4 a b+b^2\right ) \sec ^4(e+f x)-8 (a-b) b \sec ^6(e+f x)-3 b^2 \sec ^8(e+f x)\right )}{6 f (a+2 b+a \cos (2 e+2 f x))^2} \]

[In]

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

[Out]

-1/6*(Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*(24*a^2*Log[Cos[e + f*x]] + 24*a*(a - b)*Sec[e + f*x]^2 - 6*(a^2
- 4*a*b + b^2)*Sec[e + f*x]^4 - 8*(a - b)*b*Sec[e + f*x]^6 - 3*b^2*Sec[e + f*x]^8))/(f*(a + 2*b + a*Cos[2*e +
2*f*x])^2)

Maple [A] (verified)

Time = 9.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.86

method	result	size
parts	\(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{8}}{8}+\frac {\tan \left (f x +e \right )^{6}}{6}\right )}{f}+\frac {a b \tan \left (f x +e \right )^{6}}{3 f}\)	\(86\)
derivativedivides	\(\frac {\frac {b^{2} \sec \left (f x +e \right )^{8}}{8}+\frac {b \sec \left (f x +e \right )^{6} a}{3}-\frac {b^{2} \sec \left (f x +e \right )^{6}}{3}+\frac {a^{2} \sec \left (f x +e \right )^{4}}{4}-a b \sec \left (f x +e \right )^{4}+\frac {\sec \left (f x +e \right )^{4} b^{2}}{4}-a^{2} \sec \left (f x +e \right )^{2}+\sec \left (f x +e \right )^{2} a b +a^{2} \ln \left (\sec \left (f x +e \right )\right )}{f}\)	\(117\)
default	\(\frac {\frac {b^{2} \sec \left (f x +e \right )^{8}}{8}+\frac {b \sec \left (f x +e \right )^{6} a}{3}-\frac {b^{2} \sec \left (f x +e \right )^{6}}{3}+\frac {a^{2} \sec \left (f x +e \right )^{4}}{4}-a b \sec \left (f x +e \right )^{4}+\frac {\sec \left (f x +e \right )^{4} b^{2}}{4}-a^{2} \sec \left (f x +e \right )^{2}+\sec \left (f x +e \right )^{2} a b +a^{2} \ln \left (\sec \left (f x +e \right )\right )}{f}\)	\(117\)
risch	\(i a^{2} x +\frac {2 i a^{2} e}{f}+\frac {-4 a^{2} {\mathrm e}^{14 i \left (f x +e \right )}+4 a b \,{\mathrm e}^{14 i \left (f x +e \right )}-20 a^{2} {\mathrm e}^{12 i \left (f x +e \right )}+8 a b \,{\mathrm e}^{12 i \left (f x +e \right )}+4 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}-44 a^{2} {\mathrm e}^{10 i \left (f x +e \right )}+\frac {52 a b \,{\mathrm e}^{10 i \left (f x +e \right )}}{3}-\frac {16 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}}{3}-56 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+\frac {80 a b \,{\mathrm e}^{8 i \left (f x +e \right )}}{3}+\frac {40 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}}{3}-44 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+\frac {52 a b \,{\mathrm e}^{6 i \left (f x +e \right )}}{3}-\frac {16 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}}{3}-20 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+8 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+4 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-4 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+4 a b \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{8}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}\)	\(317\)

[In]

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

[Out]

a^2/f*(1/4*tan(f*x+e)^4-1/2*tan(f*x+e)^2+1/2*ln(1+tan(f*x+e)^2))+b^2/f*(1/8*tan(f*x+e)^8+1/6*tan(f*x+e)^6)+1/3
*a*b/f*tan(f*x+e)^6

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=-\frac {24 \, a^{2} \cos \left (f x + e\right )^{8} \log \left (-\cos \left (f x + e\right )\right ) + 24 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{6} - 6 \, {\left (a^{2} - 4 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, b^{2}}{24 \, f \cos \left (f x + e\right )^{8}} \]

[In]

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

[Out]

-1/24*(24*a^2*cos(f*x + e)^8*log(-cos(f*x + e)) + 24*(a^2 - a*b)*cos(f*x + e)^6 - 6*(a^2 - 4*a*b + b^2)*cos(f*
x + e)^4 - 8*(a*b - b^2)*cos(f*x + e)^2 - 3*b^2)/(f*cos(f*x + e)^8)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (85) = 170\).

Time = 1.13 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.90 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=\begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a^{2} \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {a^{2} \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac {a b \tan ^{4}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{3 f} - \frac {a b \tan ^{2}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{3 f} + \frac {a b \sec ^{2}{\left (e + f x \right )}}{3 f} + \frac {b^{2} \tan ^{4}{\left (e + f x \right )} \sec ^{4}{\left (e + f x \right )}}{8 f} - \frac {b^{2} \tan ^{2}{\left (e + f x \right )} \sec ^{4}{\left (e + f x \right )}}{12 f} + \frac {b^{2} \sec ^{4}{\left (e + f x \right )}}{24 f} & \text {for}\: f \neq 0 \\x \left (a + b \sec ^{2}{\left (e \right )}\right )^{2} \tan ^{5}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*log(tan(e + f*x)**2 + 1)/(2*f) + a**2*tan(e + f*x)**4/(4*f) - a**2*tan(e + f*x)**2/(2*f) + a*b
*tan(e + f*x)**4*sec(e + f*x)**2/(3*f) - a*b*tan(e + f*x)**2*sec(e + f*x)**2/(3*f) + a*b*sec(e + f*x)**2/(3*f)
+ b**2*tan(e + f*x)**4*sec(e + f*x)**4/(8*f) - b**2*tan(e + f*x)**2*sec(e + f*x)**4/(12*f) + b**2*sec(e + f*x
)**4/(24*f), Ne(f, 0)), (x*(a + b*sec(e)**2)**2*tan(e)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.47 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=-\frac {12 \, a^{2} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac {24 \, {\left (a^{2} - a b\right )} \sin \left (f x + e\right )^{6} - 6 \, {\left (11 \, a^{2} - 8 \, a b - b^{2}\right )} \sin \left (f x + e\right )^{4} + 4 \, {\left (15 \, a^{2} - 8 \, a b - b^{2}\right )} \sin \left (f x + e\right )^{2} - 18 \, a^{2} + 8 \, a b + b^{2}}{\sin \left (f x + e\right )^{8} - 4 \, \sin \left (f x + e\right )^{6} + 6 \, \sin \left (f x + e\right )^{4} - 4 \, \sin \left (f x + e\right )^{2} + 1}}{24 \, f} \]

[In]

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

[Out]

-1/24*(12*a^2*log(sin(f*x + e)^2 - 1) - (24*(a^2 - a*b)*sin(f*x + e)^6 - 6*(11*a^2 - 8*a*b - b^2)*sin(f*x + e)
^4 + 4*(15*a^2 - 8*a*b - b^2)*sin(f*x + e)^2 - 18*a^2 + 8*a*b + b^2)/(sin(f*x + e)^8 - 4*sin(f*x + e)^6 + 6*si
n(f*x + e)^4 - 4*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. 431 vs. \(2 (94) = 188\).

Time = 1.89 (sec) , antiderivative size = 431, normalized size of antiderivative = 4.31 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=\frac {12 \, a^{2} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right ) - 12 \, a^{2} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right ) + \frac {25 \, a^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{4} + 248 \, a^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{3} + 984 \, a^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{2} + 1760 \, a^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 512 \, a b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 256 \, b^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 1168 \, a^{2} - 1024 \, a b + 256 \, b^{2}}{{\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}^{4}}}{24 \, f} \]

[In]

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

[Out]

1/24*(12*a^2*log(abs(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2)) - 12
*a^2*log(abs(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2)) + (25*a^2*((
cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^4 + 248*a^2*((cos(f*x + e) + 1)/
(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^3 + 984*a^2*((cos(f*x + e) + 1)/(cos(f*x + e) - 1)
+ (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^2 + 1760*a^2*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e)
- 1)/(cos(f*x + e) + 1)) - 512*a*b*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) +
1)) - 256*b^2*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) + 1168*a^2 - 10
24*a*b + 256*b^2)/((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2)^4)/f

Mupad [B] (verification not implemented)

Time = 19.52 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.24 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^5(e+f x) \, dx=\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {{\left (a+b\right )}^2}{4}+\frac {b^2}{4}-\frac {b\,\left (a+b\right )}{2}\right )}{f}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {{\left (a+b\right )}^2}{2}+\frac {b^2}{2}-b\,\left (a+b\right )\right )}{f}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (\frac {b^2}{6}-\frac {b\,\left (a+b\right )}{3}\right )}{f}+\frac {a^2\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^8}{8\,f} \]

[In]

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

[Out]

(tan(e + f*x)^4*((a + b)^2/4 + b^2/4 - (b*(a + b))/2))/f - (tan(e + f*x)^2*((a + b)^2/2 + b^2/2 - b*(a + b)))/
f - (tan(e + f*x)^6*(b^2/6 - (b*(a + b))/3))/f + (a^2*log(tan(e + f*x)^2 + 1))/(2*f) + (b^2*tan(e + f*x)^8)/(8
*f)