3.254 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^2}{a+b \sec (e+f x)} \, dx\)

Optimal. Leaf size=103 \[ \frac{d (2 b c-a d) \tanh ^{-1}(\sin (e+f x))}{b^2 f}+\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b^2 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^2 \tan (e+f x)}{b f} \]

[Out]

(d*(2*b*c - a*d)*ArcTanh[Sin[e + f*x]])/(b^2*f) + (2*(b*c - a*d)^2*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt
[a + b]])/(Sqrt[a - b]*b^2*Sqrt[a + b]*f) + (d^2*Tan[e + f*x])/(b*f)

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Rubi [A]  time = 0.297448, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3988, 2952, 2659, 208, 3770, 3767, 8} \[ \frac{d (2 b c-a d) \tanh ^{-1}(\sin (e+f x))}{b^2 f}+\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b^2 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^2 \tan (e+f x)}{b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*b*c - a*d)*ArcTanh[Sin[e + f*x]])/(b^2*f) + (2*(b*c - a*d)^2*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt
[a + b]])/(Sqrt[a - b]*b^2*Sqrt[a + b]*f) + (d^2*Tan[e + f*x])/(b*f)

Rule 3988

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[1/g^(m + n), Int[(g*Csc[e + f*x])^(m + n + p)*(b + a*Sin[e + f*x])^m*(d
+ c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && IntegerQ[m] && Inte
gerQ[n]

Rule 2952

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (IntegersQ[m, n] || IntegersQ[m, p
] || IntegersQ[n, p]) && NeQ[p, 2]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3770

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^2}{a+b \sec (e+f x)} \, dx &=\int \frac{(d+c \cos (e+f x))^2 \sec ^2(e+f x)}{b+a \cos (e+f x)} \, dx\\ &=\int \left (\frac{(b c-a d)^2}{b^2 (b+a \cos (e+f x))}+\frac{d (2 b c-a d) \sec (e+f x)}{b^2}+\frac{d^2 \sec ^2(e+f x)}{b}\right ) \, dx\\ &=\frac{d^2 \int \sec ^2(e+f x) \, dx}{b}+\frac{(b c-a d)^2 \int \frac{1}{b+a \cos (e+f x)} \, dx}{b^2}+\frac{(d (2 b c-a d)) \int \sec (e+f x) \, dx}{b^2}\\ &=\frac{d (2 b c-a d) \tanh ^{-1}(\sin (e+f x))}{b^2 f}-\frac{d^2 \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{b f}+\frac{\left (2 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{b^2 f}\\ &=\frac{d (2 b c-a d) \tanh ^{-1}(\sin (e+f x))}{b^2 f}+\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^2 \sqrt{a+b} f}+\frac{d^2 \tan (e+f x)}{b f}\\ \end{align*}

Mathematica [A]  time = 0.804488, size = 135, normalized size = 1.31 \[ \frac{d \left (b d \tan (e+f x)-(2 b c-a d) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )\right )-\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(b*c - a*d)^2*ArcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + d*(-((2*b*c - a*d)*
(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])) + b*d*Tan[e + f*x]))/(b
^2*f)

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Maple [B]  time = 0.065, size = 288, normalized size = 2.8 \begin{align*} 2\,{\frac{{a}^{2}{d}^{2}}{f{b}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{acd}{fb\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{{c}^{2}}{f\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{{d}^{2}}{fb} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a{d}^{2}}{f{b}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }+2\,{\frac{d\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) c}{fb}}-{\frac{{d}^{2}}{fb} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a{d}^{2}}{f{b}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) }-2\,{\frac{d\ln \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) c}{fb}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+b*sec(f*x+e)),x)

[Out]

2/f/b^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a^2*d^2-4/f/b/((a+b)*(a-b))^
(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a*c*d+2/f/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/
2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*c^2-1/f*d^2/b/(tan(1/2*f*x+1/2*e)+1)-1/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)+1)*a+
2/f*d/b*ln(tan(1/2*f*x+1/2*e)+1)*c-1/f*d^2/b/(tan(1/2*f*x+1/2*e)-1)+1/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)-1)*a-2/f
*d/b*ln(tan(1/2*f*x+1/2*e)-1)*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 12.7316, size = 1139, normalized size = 11.06 \begin{align*} \left [\frac{2 \,{\left (a^{2} b - b^{3}\right )} d^{2} \sin \left (f x + e\right ) +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{a^{2} - b^{2}} \cos \left (f x + e\right ) \log \left (\frac{2 \, a b \cos \left (f x + e\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (f x + e\right )^{2} + 2 \, a b \cos \left (f x + e\right ) + b^{2}}\right ) +{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} f \cos \left (f x + e\right )}, \frac{2 \,{\left (a^{2} b - b^{3}\right )} d^{2} \sin \left (f x + e\right ) + 2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (f x + e\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) +{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (\sin \left (f x + e\right ) + 1\right ) -{\left (2 \,{\left (a^{2} b - b^{3}\right )} c d -{\left (a^{3} - a b^{2}\right )} d^{2}\right )} \cos \left (f x + e\right ) \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} f \cos \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(a^2*b - b^3)*d^2*sin(f*x + e) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(a^2 - b^2)*cos(f*x + e)*log((2*a
*b*cos(f*x + e) - (a^2 - 2*b^2)*cos(f*x + e)^2 + 2*sqrt(a^2 - b^2)*(b*cos(f*x + e) + a)*sin(f*x + e) + 2*a^2 -
b^2)/(a^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + b^2)) + (2*(a^2*b - b^3)*c*d - (a^3 - a*b^2)*d^2)*cos(f*x + e
)*log(sin(f*x + e) + 1) - (2*(a^2*b - b^3)*c*d - (a^3 - a*b^2)*d^2)*cos(f*x + e)*log(-sin(f*x + e) + 1))/((a^2
*b^2 - b^4)*f*cos(f*x + e)), 1/2*(2*(a^2*b - b^3)*d^2*sin(f*x + e) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-a
^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(f*x + e) + a)/((a^2 - b^2)*sin(f*x + e)))*cos(f*x + e) + (2*(a^2*b -
b^3)*c*d - (a^3 - a*b^2)*d^2)*cos(f*x + e)*log(sin(f*x + e) + 1) - (2*(a^2*b - b^3)*c*d - (a^3 - a*b^2)*d^2)*
cos(f*x + e)*log(-sin(f*x + e) + 1))/((a^2*b^2 - b^4)*f*cos(f*x + e))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**2/(a+b*sec(f*x+e)),x)

[Out]

Integral((c + d*sec(e + f*x))**2*sec(e + f*x)/(a + b*sec(e + f*x)), x)

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Giac [B]  time = 1.41525, size = 273, normalized size = 2.65 \begin{align*} -\frac{\frac{2 \, d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )} b} - \frac{{\left (2 \, b c d - a d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{b^{2}} + \frac{{\left (2 \, b c d - a d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{b^{2}} + \frac{2 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{\sqrt{-a^{2} + b^{2}} b^{2}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*d^2*tan(1/2*f*x + 1/2*e)/((tan(1/2*f*x + 1/2*e)^2 - 1)*b) - (2*b*c*d - a*d^2)*log(abs(tan(1/2*f*x + 1/2*e)
+ 1))/b^2 + (2*b*c*d - a*d^2)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(pi*
floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*f*x + 1/2*e) - b*tan(1/2*f*x + 1/2*e))/sqrt(-
a^2 + b^2)))/(sqrt(-a^2 + b^2)*b^2))/f