3.363 \(\int \frac {\tan ^5(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=78 \[ \frac {(a+b)^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac {a+b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac {\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

[Out]

1/4*(a+b)^2/a^3/f/(b+a*cos(f*x+e)^2)^2+(-a-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 = 0.11, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4138, 444, 43} \[ \frac {(a+b)^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac {a+b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac {\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(a + b)^2/(4*a^3*f*(b + a*Cos[e + f*x]^2)^2) - (a + 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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 ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x \left (1-x^2\right )^2}{\left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(1-x)^2}{(b+a x)^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {(a+b)^2}{a^2 (b+a x)^3}-\frac {2 (a+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 {(a+b)^2}{4 a^3 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {a+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 = 2.15, size = 136, normalized size = 1.74 \[ -\frac {2 \left (a^2+4 a b+3 b^2\right )+a^2 \cos ^2(2 (e+f x)) \log (a \cos (2 (e+f x))+a+2 b)+(a+2 b)^2 \log (a \cos (2 (e+f x))+a+2 b)+2 a \cos (2 (e+f x)) ((a+2 b) \log (a \cos (2 (e+f x))+a+2 b)+2 (a+b))}{2 a^3 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

-1/2*(2*(a^2 + 4*a*b + 3*b^2) + (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*(a + 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 = 0.54, size = 116, normalized size = 1.49 \[ -\frac {4 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} - a^{2} + 2 \, a b + 3 \, b^{2} + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right )}{4 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

-1/4*(4*(a^2 + a*b)*cos(f*x + e)^2 - a^2 + 2*a*b + 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)^5/(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^2+6*((1-cos(f*x+e
xp(1)))/(1+cos(f*x+exp(1))))^4*b*a+3*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^4*a^2+12*((1-cos(f*x+exp(1)))/(
1+cos(f*x+exp(1))))^3*b^2-8*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*b*a-20*((1-cos(f*x+exp(1)))/(1+cos(f*x
+exp(1))))^3*a^2+18*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2-28*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))
))^2*b*a+50*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^2+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b^2-8*(
1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b*a-20*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^2+3*b^2+6*b*a+3*a^2)*1
/8/a^3/(((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 = 0.86, size = 138, normalized size = 1.77 \[ \frac {1}{4 f a \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b}{2 f \,a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{2}}{4 a^{3} f \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {\ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{2 a^{3} f}-\frac {1}{f \,a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}-\frac {b}{a^{3} f \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x)

[Out]

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

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maxima [A]  time = 0.35, size = 112, normalized size = 1.44 \[ \frac {\frac {4 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{2} - 3 \, a^{2} - 6 \, a b - 3 \, b^{2}}{a^{5} \sin \left (f x + e\right )^{4} + a^{5} + 2 \, a^{4} b + a^{3} b^{2} - 2 \, {\left (a^{5} + a^{4} b\right )} \sin \left (f x + e\right )^{2}} - \frac {2 \, \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3}}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.60, size = 166, normalized size = 2.13 \[ \frac {\mathrm {atanh}\left (\frac {4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,b^2+\frac {8\,b^3}{a}+4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {8\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{a}}\right )}{a^3\,f}+\frac {\frac {-a^3+3\,a\,b^2+2\,b^3}{4\,a^2\,b^2}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a^2-b^2\right )}{2\,a^2\,b}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^5/(a + b/cos(e + f*x)^2)^3,x)

[Out]

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

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

[Out]

Timed out

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