3.365 \(\int \frac {\tan (e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=74 \[ \frac {b^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac {b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac {\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 74, 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} \[ \frac {b^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac {b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac {\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

b^2/(4*a^3*f*(b + a*Cos[e + f*x]^2)^2) - b/(a^3*f*(b + a*Cos[e + f*x]^2)) - Log[b + a*Cos[e + f*x]^2]/(2*a^3*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 \frac {\tan (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^5}{\left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {x^2}{(b+a x)^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {b^2}{a^2 (b+a x)^3}-\frac {2 b}{a^2 (b+a x)^2}+\frac {1}{a^2 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac {b^2}{4 a^3 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b}{a^3 f \left (b+a \cos ^2(e+f x)\right )}-\frac {\log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 f}\\ \end {align*}

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Mathematica [A]  time = 1.33, size = 129, normalized size = 1.74 \[ -\frac {a^2 \cos ^2(2 (e+f x)) \log (a \cos (2 (e+f x))+a+2 b)+(a+2 b)^2 \log (a \cos (2 (e+f x))+a+2 b)+2 a \cos (2 (e+f x)) ((a+2 b) \log (a \cos (2 (e+f x))+a+2 b)+2 b)+2 b (2 a+3 b)}{2 a^3 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

-1/2*(2*b*(2*a + 3*b) + (a + 2*b)^2*Log[a + 2*b + a*Cos[2*(e + f*x)]] + a^2*Cos[2*(e + f*x)]^2*Log[a + 2*b + a
*Cos[2*(e + f*x)]] + 2*a*Cos[2*(e + f*x)]*(2*b + (a + 2*b)*Log[a + 2*b + a*Cos[2*(e + f*x)]]))/(a^3*f*(a + 2*b
+ a*Cos[2*(e + f*x)])^2)

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fricas [A]  time = 0.53, size = 102, normalized size = 1.38 \[ -\frac {4 \, a b \cos \left (f x + e\right )^{2} + 3 \, b^{2} + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right )}{4 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

-1/4*(4*a*b*cos(f*x + e)^2 + 3*b^2 + 2*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*log(a*cos(f*x + e)^2
+ b))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f)

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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(tan(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)2/f*(1/2/a^3*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1))-1/4/a^3*ln(((1-cos(f*x+exp(1)))/(1+cos(f*x
+exp(1))))^2*b+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b-2*(1-
cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+b+a)+(3*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*b^4+12*((1-cos(f*x+
exp(1)))/(1+cos(f*x+exp(1))))^4*b^3*a+18*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*b^2*a^2+12*((1-cos(f*x+ex
p(1)))/(1+cos(f*x+exp(1))))^4*b*a^3+3*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^4+12*((1-cos(f*x+exp(1)))/
(1+cos(f*x+exp(1))))^3*b^4+16*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*b^3*a-24*((1-cos(f*x+exp(1)))/(1+cos
(f*x+exp(1))))^3*b^2*a^2-40*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*b*a^3-12*((1-cos(f*x+exp(1)))/(1+cos(f
*x+exp(1))))^3*a^4+18*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^4+8*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)
)))^2*b^3*a+12*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2*a^2+56*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))
)^2*b*a^3+18*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^4+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^4+16
*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^3*a-24*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2*a^2-40*(1-cos(f*
x+exp(1)))/(1+cos(f*x+exp(1)))*b*a^3-12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^4+3*b^4+12*b^3*a+18*b^2*a^2+
12*b*a^3+3*a^4)/(8*b^2*a^3+16*b*a^4+8*a^5)/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b+((1-cos(f*x+exp(1)))
/(1+cos(f*x+exp(1))))^2*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))
)*a+b+a)^2)

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maple [A]  time = 0.39, size = 81, normalized size = 1.09 \[ \frac {1}{4 f a \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {\ln \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )}{2 f \,a^{3}}+\frac {1}{2 f \,a^{2} \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )}+\frac {\ln \left (\sec \left (f x +e \right )\right )}{f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)/(a+b*sec(f*x+e)^2)^3,x)

[Out]

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

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maxima [A]  time = 0.37, size = 102, normalized size = 1.38 \[ \frac {\frac {4 \, a b \sin \left (f x + e\right )^{2} - 4 \, a b - 3 \, b^{2}}{a^{5} \sin \left (f x + e\right )^{4} + a^{5} + 2 \, a^{4} b + a^{3} b^{2} - 2 \, {\left (a^{5} + a^{4} b\right )} \sin \left (f x + e\right )^{2}} - \frac {2 \, \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/4*((4*a*b*sin(f*x + e)^2 - 4*a*b - 3*b^2)/(a^5*sin(f*x + e)^4 + a^5 + 2*a^4*b + a^3*b^2 - 2*(a^5 + a^4*b)*si
n(f*x + e)^2) - 2*log(a*sin(f*x + e)^2 - a - b)/a^3)/f

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mupad [B]  time = 4.40, size = 142, normalized size = 1.92 \[ \frac {\frac {3\,a+2\,b}{4\,a^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,a^2}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}+\frac {\mathrm {atanh}\left (\frac {4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,b^2+\frac {8\,b^3}{a}+4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {8\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{a}}\right )}{a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)/(a + b/cos(e + f*x)^2)^3,x)

[Out]

((3*a + 2*b)/(4*a^2) + (b*tan(e + f*x)^2)/(2*a^2))/(f*(2*a*b + a^2 + b^2 + tan(e + f*x)^2*(2*a*b + 2*b^2) + b^
2*tan(e + f*x)^4)) + atanh((4*b^2*tan(e + f*x)^2)/(8*b^2 + (8*b^3)/a + 4*b^2*tan(e + f*x)^2 + (8*b^3*tan(e + f
*x)^2)/a))/(a^3*f)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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