3.153 \(\int \sec ^3(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=70 \[ \frac{(4 a+3 b) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{(4 a+3 b) \tan (e+f x) \sec (e+f x)}{8 f}+\frac{b \tan (e+f x) \sec ^3(e+f x)}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.0457003, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4046, 3768, 3770} \[ \frac{(4 a+3 b) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{(4 a+3 b) \tan (e+f x) \sec (e+f x)}{8 f}+\frac{b \tan (e+f x) \sec ^3(e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a + 3*b)*ArcTanh[Sin[e + f*x]])/(8*f) + ((4*a + 3*b)*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (b*Sec[e + f*x]^3*
Tan[e + f*x])/(4*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 ^3(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{b \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac{1}{4} (4 a+3 b) \int \sec ^3(e+f x) \, dx\\ &=\frac{(4 a+3 b) \sec (e+f x) \tan (e+f x)}{8 f}+\frac{b \sec ^3(e+f x) \tan (e+f x)}{4 f}+\frac{1}{8} (4 a+3 b) \int \sec (e+f x) \, dx\\ &=\frac{(4 a+3 b) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{(4 a+3 b) \sec (e+f x) \tan (e+f x)}{8 f}+\frac{b \sec ^3(e+f x) \tan (e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.123698, size = 54, normalized size = 0.77 \[ \frac{(4 a+3 b) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \sec (e+f x) \left (4 a+2 b \sec ^2(e+f x)+3 b\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.028, size = 98, normalized size = 1.4 \begin{align*}{\frac{a\tan \left ( fx+e \right ) \sec \left ( fx+e \right ) }{2\,f}}+{\frac{a\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+{\frac{b \left ( \sec \left ( fx+e \right ) \right ) ^{3}\tan \left ( fx+e \right ) }{4\,f}}+{\frac{3\,b\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{3\,b\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.994445, size = 131, normalized size = 1.87 \begin{align*} \frac{{\left (4 \, a + 3 \, b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (4 \, a + 3 \, b\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac{2 \,{\left ({\left (4 \, a + 3 \, b\right )} \sin \left (f x + e\right )^{3} -{\left (4 \, a + 5 \, b\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1}}{16 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*a + 3*b)*log(sin(f*x + e) + 1) - (4*a + 3*b)*log(sin(f*x + e) - 1) - 2*((4*a + 3*b)*sin(f*x + e)^3 -
(4*a + 5*b)*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1))/f

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Fricas [A]  time = 0.507906, size = 243, normalized size = 3.47 \begin{align*} \frac{{\left (4 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (4 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left ({\left (4 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{2} + 2 \, b\right )} \sin \left (f x + e\right )}{16 \, f \cos \left (f x + e\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*a + 3*b)*cos(f*x + e)^4*log(sin(f*x + e) + 1) - (4*a + 3*b)*cos(f*x + e)^4*log(-sin(f*x + e) + 1) + 2
*((4*a + 3*b)*cos(f*x + e)^2 + 2*b)*sin(f*x + e))/(f*cos(f*x + e)^4)

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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 ^{3}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29215, size = 142, normalized size = 2.03 \begin{align*} \frac{{\left (4 \, a + 3 \, b\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (4 \, a + 3 \, b\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - \frac{2 \,{\left (4 \, a \sin \left (f x + e\right )^{3} + 3 \, b \sin \left (f x + e\right )^{3} - 4 \, a \sin \left (f x + e\right ) - 5 \, b \sin \left (f x + e\right )\right )}}{{\left (\sin \left (f x + e\right )^{2} - 1\right )}^{2}}}{16 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*a + 3*b)*log(sin(f*x + e) + 1) - (4*a + 3*b)*log(-sin(f*x + e) + 1) - 2*(4*a*sin(f*x + e)^3 + 3*b*sin
(f*x + e)^3 - 4*a*sin(f*x + e) - 5*b*sin(f*x + e))/(sin(f*x + e)^2 - 1)^2)/f