3.152 \(\int \sec ^5(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=98 \[ \frac{(6 a+5 b) \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac{(6 a+5 b) \tan (e+f x) \sec ^3(e+f x)}{24 f}+\frac{(6 a+5 b) \tan (e+f x) \sec (e+f x)}{16 f}+\frac{b \tan (e+f x) \sec ^5(e+f x)}{6 f} \]

[Out]

((6*a + 5*b)*ArcTanh[Sin[e + f*x]])/(16*f) + ((6*a + 5*b)*Sec[e + f*x]*Tan[e + f*x])/(16*f) + ((6*a + 5*b)*Sec
[e + f*x]^3*Tan[e + f*x])/(24*f) + (b*Sec[e + f*x]^5*Tan[e + f*x])/(6*f)

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Rubi [A]  time = 0.0593756, antiderivative size = 98, normalized size of antiderivative = 1., 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 = {4046, 3768, 3770} \[ \frac{(6 a+5 b) \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac{(6 a+5 b) \tan (e+f x) \sec ^3(e+f x)}{24 f}+\frac{(6 a+5 b) \tan (e+f x) \sec (e+f x)}{16 f}+\frac{b \tan (e+f x) \sec ^5(e+f x)}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a + 5*b)*ArcTanh[Sin[e + f*x]])/(16*f) + ((6*a + 5*b)*Sec[e + f*x]*Tan[e + f*x])/(16*f) + ((6*a + 5*b)*Sec
[e + f*x]^3*Tan[e + f*x])/(24*f) + (b*Sec[e + f*x]^5*Tan[e + f*x])/(6*f)

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sec ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{b \sec ^5(e+f x) \tan (e+f x)}{6 f}+\frac{1}{6} (6 a+5 b) \int \sec ^5(e+f x) \, dx\\ &=\frac{(6 a+5 b) \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac{b \sec ^5(e+f x) \tan (e+f x)}{6 f}+\frac{1}{8} (6 a+5 b) \int \sec ^3(e+f x) \, dx\\ &=\frac{(6 a+5 b) \sec (e+f x) \tan (e+f x)}{16 f}+\frac{(6 a+5 b) \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac{b \sec ^5(e+f x) \tan (e+f x)}{6 f}+\frac{1}{16} (6 a+5 b) \int \sec (e+f x) \, dx\\ &=\frac{(6 a+5 b) \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac{(6 a+5 b) \sec (e+f x) \tan (e+f x)}{16 f}+\frac{(6 a+5 b) \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac{b \sec ^5(e+f x) \tan (e+f x)}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.313931, size = 75, normalized size = 0.77 \[ \frac{3 (6 a+5 b) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \sec (e+f x) \left (2 (6 a+5 b) \sec ^2(e+f x)+3 (6 a+5 b)+8 b \sec ^4(e+f x)\right )}{48 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(6*a + 5*b)*ArcTanh[Sin[e + f*x]] + Sec[e + f*x]*(3*(6*a + 5*b) + 2*(6*a + 5*b)*Sec[e + f*x]^2 + 8*b*Sec[e
+ f*x]^4)*Tan[e + f*x])/(48*f)

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Maple [A]  time = 0.031, size = 138, normalized size = 1.4 \begin{align*}{\frac{a\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{3\,a\tan \left ( fx+e \right ) \sec \left ( fx+e \right ) }{8\,f}}+{\frac{3\,a\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{b \left ( \sec \left ( fx+e \right ) \right ) ^{5}\tan \left ( fx+e \right ) }{6\,f}}+{\frac{5\,b \left ( \sec \left ( fx+e \right ) \right ) ^{3}\tan \left ( fx+e \right ) }{24\,f}}+{\frac{5\,b\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{16\,f}}+{\frac{5\,b\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{16\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*a*tan(f*x+e)*sec(f*x+e)^3+3/8/f*a*tan(f*x+e)*sec(f*x+e)+3/8/f*a*ln(sec(f*x+e)+tan(f*x+e))+1/6*b*sec(f*x+
e)^5*tan(f*x+e)/f+5/24*b*sec(f*x+e)^3*tan(f*x+e)/f+5/16*b*sec(f*x+e)*tan(f*x+e)/f+5/16/f*b*ln(sec(f*x+e)+tan(f
*x+e))

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Maxima [A]  time = 0.99761, size = 170, normalized size = 1.73 \begin{align*} \frac{3 \,{\left (6 \, a + 5 \, b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (6 \, a + 5 \, b\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac{2 \,{\left (3 \,{\left (6 \, a + 5 \, b\right )} \sin \left (f x + e\right )^{5} - 8 \,{\left (6 \, a + 5 \, b\right )} \sin \left (f x + e\right )^{3} + 3 \,{\left (10 \, a + 11 \, b\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(6*a + 5*b)*log(sin(f*x + e) + 1) - 3*(6*a + 5*b)*log(sin(f*x + e) - 1) - 2*(3*(6*a + 5*b)*sin(f*x + e
)^5 - 8*(6*a + 5*b)*sin(f*x + e)^3 + 3*(10*a + 11*b)*sin(f*x + e))/(sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(
f*x + e)^2 - 1))/f

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Fricas [A]  time = 0.517976, size = 293, normalized size = 2.99 \begin{align*} \frac{3 \,{\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (3 \,{\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (6 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{2} + 8 \, b\right )} \sin \left (f x + e\right )}{96 \, f \cos \left (f x + e\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(6*a + 5*b)*cos(f*x + e)^6*log(sin(f*x + e) + 1) - 3*(6*a + 5*b)*cos(f*x + e)^6*log(-sin(f*x + e) + 1)
+ 2*(3*(6*a + 5*b)*cos(f*x + e)^4 + 2*(6*a + 5*b)*cos(f*x + e)^2 + 8*b)*sin(f*x + e))/(f*cos(f*x + e)^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sec ^{5}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32571, size = 176, normalized size = 1.8 \begin{align*} \frac{3 \,{\left (6 \, a + 5 \, b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (6 \, a + 5 \, b\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - \frac{2 \,{\left (18 \, a \sin \left (f x + e\right )^{5} + 15 \, b \sin \left (f x + e\right )^{5} - 48 \, a \sin \left (f x + e\right )^{3} - 40 \, b \sin \left (f x + e\right )^{3} + 30 \, a \sin \left (f x + e\right ) + 33 \, b \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{3}}}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(6*a + 5*b)*log(sin(f*x + e) + 1) - 3*(6*a + 5*b)*log(-sin(f*x + e) + 1) - 2*(18*a*sin(f*x + e)^5 + 15
*b*sin(f*x + e)^5 - 48*a*sin(f*x + e)^3 - 40*b*sin(f*x + e)^3 + 30*a*sin(f*x + e) + 33*b*sin(f*x + e))/(sin(f*
x + e)^2 - 1)^3)/f