3.326 \(\int (a+b \sec ^2(e+f x))^2 \tan (e+f x) \, dx\)

Optimal. Leaf size=48 \[ -\frac {a^2 \log (\cos (e+f x))}{f}+\frac {a b \sec ^2(e+f x)}{f}+\frac {b^2 \sec ^4(e+f x)}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4138, 266, 43} \[ -\frac {a^2 \log (\cos (e+f x))}{f}+\frac {a b \sec ^2(e+f x)}{f}+\frac {b^2 \sec ^4(e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rule 4138

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*} \int \left (a+b \sec ^2(e+f x)\right )^2 \tan (e+f x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (b+a x^2\right )^2}{x^5} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^2}{x^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {b^2}{x^3}+\frac {2 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 b \sec ^2(e+f x)}{f}+\frac {b^2 \sec ^4(e+f x)}{4 f}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 82, normalized size = 1.71 \[ -\frac {\sec ^4(e+f x) \left (a \cos ^2(e+f x)+b\right )^2 \left (4 a^2 \cos ^4(e+f x) \log (\cos (e+f x))-4 a b \cos ^2(e+f x)-b^2\right )}{f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 53, normalized size = 1.10 \[ -\frac {4 \, a^{2} \cos \left (f x + e\right )^{4} \log \left (-\cos \left (f x + e\right )\right ) - 4 \, a b \cos \left (f x + e\right )^{2} - b^{2}}{4 \, f \cos \left (f x + e\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*(a^2/4*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1
-cos(f*x+exp(1)))*(1+cos(f*x+exp(1)))+2))-a^2/4*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+ex
p(1)))*(1+cos(f*x+exp(1)))-2))+(3*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+ex
p(1))))^2*a^2-12*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*a^2+16*((
1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*a*b+8*((1-cos(f*x+exp(1)))/(
1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*b^2+12*a^2-32*a*b)*1/8/((1-cos(f*x+exp(1)))/(1+c
os(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1)))-2)^2)

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maple [A]  time = 0.34, size = 46, normalized size = 0.96 \[ \frac {b^{2} \left (\sec ^{4}\left (f x +e \right )\right )}{4 f}+\frac {a b \left (\sec ^{2}\left (f x +e \right )\right )}{f}+\frac {a^{2} \ln \left (\sec \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 67, normalized size = 1.40 \[ -\frac {2 \, a^{2} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) + \frac {4 \, a b \sin \left (f x + e\right )^{2} - 4 \, a b - b^{2}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.50, size = 61, normalized size = 1.27 \[ \frac {a^2\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {b^2}{2}-b\,\left (a+b\right )\right )}{f}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.67, size = 61, normalized size = 1.27 \[ \begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a b \sec ^{2}{\left (e + f x \right )}}{f} + \frac {b^{2} \sec ^{4}{\left (e + f x \right )}}{4 f} & \text {for}\: f \neq 0 \\x \left (a + b \sec ^{2}{\relax (e )}\right )^{2} \tan {\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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