3.253 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^3}{a+b \sec (e+f x)} \, dx\)

Optimal. Leaf size=170 \[ \frac{d \left (a^2 d^2-3 a b c d+3 b^2 c^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^3 f}+\frac{d^2 (3 b c-a d) \tan (e+f x)}{b^2 f}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b^3 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^3 \tanh ^{-1}(\sin (e+f x))}{2 b f}+\frac{d^3 \tan (e+f x) \sec (e+f x)}{2 b f} \]

[Out]

(d^3*ArcTanh[Sin[e + f*x]])/(2*b*f) + (d*(3*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*ArcTanh[Sin[e + f*x]])/(b^3*f) + (2
*(b*c - a*d)^3*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^3*Sqrt[a + b]*f) + (d^2*(3*
b*c - a*d)*Tan[e + f*x])/(b^2*f) + (d^3*Sec[e + f*x]*Tan[e + f*x])/(2*b*f)

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Rubi [A]  time = 0.350474, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {3988, 2952, 2659, 208, 3770, 3767, 8, 3768} \[ \frac{d \left (a^2 d^2-3 a b c d+3 b^2 c^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^3 f}+\frac{d^2 (3 b c-a d) \tan (e+f x)}{b^2 f}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b^3 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^3 \tanh ^{-1}(\sin (e+f x))}{2 b f}+\frac{d^3 \tan (e+f x) \sec (e+f x)}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*ArcTanh[Sin[e + f*x]])/(2*b*f) + (d*(3*b^2*c^2 - 3*a*b*c*d + a^2*d^2)*ArcTanh[Sin[e + f*x]])/(b^3*f) + (2
*(b*c - a*d)^3*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^3*Sqrt[a + b]*f) + (d^2*(3*
b*c - a*d)*Tan[e + f*x])/(b^2*f) + (d^3*Sec[e + f*x]*Tan[e + f*x])/(2*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]

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]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^3}{a+b \sec (e+f x)} \, dx &=\int \frac{(d+c \cos (e+f x))^3 \sec ^3(e+f x)}{b+a \cos (e+f x)} \, dx\\ &=\int \left (\frac{(b c-a d)^3}{b^3 (b+a \cos (e+f x))}+\frac{d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) \sec (e+f x)}{b^3}+\frac{d^2 (3 b c-a d) \sec ^2(e+f x)}{b^2}+\frac{d^3 \sec ^3(e+f x)}{b}\right ) \, dx\\ &=\frac{d^3 \int \sec ^3(e+f x) \, dx}{b}+\frac{(b c-a d)^3 \int \frac{1}{b+a \cos (e+f x)} \, dx}{b^3}+\frac{\left (d^2 (3 b c-a d)\right ) \int \sec ^2(e+f x) \, dx}{b^2}+\frac{\left (d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right )\right ) \int \sec (e+f x) \, dx}{b^3}\\ &=\frac{d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^3 f}+\frac{d^3 \sec (e+f x) \tan (e+f x)}{2 b f}+\frac{d^3 \int \sec (e+f x) \, dx}{2 b}+\frac{\left (2 (b c-a d)^3\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^3 f}-\frac{\left (d^2 (3 b c-a d)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{b^2 f}\\ &=\frac{d^3 \tanh ^{-1}(\sin (e+f x))}{2 b f}+\frac{d \left (3 b^2 c^2-3 a b c d+a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^3 f}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^3 \sqrt{a+b} f}+\frac{d^2 (3 b c-a d) \tan (e+f x)}{b^2 f}+\frac{d^3 \sec (e+f x) \tan (e+f x)}{2 b f}\\ \end{align*}

Mathematica [B]  time = 1.40604, size = 389, normalized size = 2.29 \[ \frac{\cos ^2(e+f x) (a \cos (e+f x)+b) (c+d \sec (e+f x))^3 \left (-2 d \left (2 a^2 d^2-6 a b c d+b^2 \left (6 c^2+d^2\right )\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+2 d \left (2 a^2 d^2-6 a b c d+b^2 \left (6 c^2+d^2\right )\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\frac{8 (a d-b c)^3 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{4 b d^2 (3 b c-a d) \sin \left (\frac{1}{2} (e+f x)\right )}{\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )}+\frac{4 b d^2 (3 b c-a d) \sin \left (\frac{1}{2} (e+f x)\right )}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}+\frac{b^2 d^3}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{b^2 d^3}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}\right )}{4 b^3 f (a+b \sec (e+f x)) (c \cos (e+f x)+d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^2*(b + a*Cos[e + f*x])*(c + d*Sec[e + f*x])^3*((8*(-(b*c) + a*d)^3*ArcTanh[((-a + b)*Tan[(e + f*
x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 2*d*(-6*a*b*c*d + 2*a^2*d^2 + b^2*(6*c^2 + d^2))*Log[Cos[(e + f*x)/
2] - Sin[(e + f*x)/2]] + 2*d*(-6*a*b*c*d + 2*a^2*d^2 + b^2*(6*c^2 + d^2))*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)
/2]] + (b^2*d^3)/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (4*b*d^2*(3*b*c - a*d)*Sin[(e + f*x)/2])/(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2]) - (b^2*d^3)/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (4*b*d^2*(3*b*c - a*d)*Sin[(
e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(4*b^3*f*(d + c*Cos[e + f*x])^3*(a + b*Sec[e + f*x]))

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Maple [B]  time = 0.081, size = 593, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^3/(a+b*sec(f*x+e)),x)

[Out]

-2/f/b^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a^3*d^3+6/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*c*d^2-6/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^2*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^3-1/2/f*d^3/b/(tan(1/2*f*x+1/2*e)+1)^2+1/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)+1)*a^2-3/f*d^2/b
^2*ln(tan(1/2*f*x+1/2*e)+1)*a*c+3/f*d/b*ln(tan(1/2*f*x+1/2*e)+1)*c^2+1/2/f*d^3/b*ln(tan(1/2*f*x+1/2*e)+1)+1/f*
d^3/b^2/(tan(1/2*f*x+1/2*e)+1)*a-3/f*d^2/b/(tan(1/2*f*x+1/2*e)+1)*c+1/2/f*d^3/b/(tan(1/2*f*x+1/2*e)+1)+1/2/f*d
^3/b/(tan(1/2*f*x+1/2*e)-1)^2-1/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)-1)*a^2+3/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)-1)*a*
c-3/f*d/b*ln(tan(1/2*f*x+1/2*e)-1)*c^2-1/2/f*d^3/b*ln(tan(1/2*f*x+1/2*e)-1)+1/f*d^3/b^2/(tan(1/2*f*x+1/2*e)-1)
*a-3/f*d^2/b/(tan(1/2*f*x+1/2*e)-1)*c+1/2/f*d^3/b/(tan(1/2*f*x+1/2*e)-1)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 85.8349, size = 1621, normalized size = 9.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(a^2 - b^2)*cos(f*x + e)^2*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)) - (6*(a^2*b^2 - b^4)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 + (2*a^4 - a
^2*b^2 - b^4)*d^3)*cos(f*x + e)^2*log(sin(f*x + e) + 1) + (6*(a^2*b^2 - b^4)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 +
(2*a^4 - a^2*b^2 - b^4)*d^3)*cos(f*x + e)^2*log(-sin(f*x + e) + 1) - 2*((a^2*b^2 - b^4)*d^3 + 2*(3*(a^2*b^2 -
b^4)*c*d^2 - (a^3*b - a*b^3)*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*b^3 - b^5)*f*cos(f*x + e)^2), 1/4*(4*(b^3
*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*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 + (6*(a^2*b^2 - b^4)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 + (2*a^4 - a^
2*b^2 - b^4)*d^3)*cos(f*x + e)^2*log(sin(f*x + e) + 1) - (6*(a^2*b^2 - b^4)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 +
(2*a^4 - a^2*b^2 - b^4)*d^3)*cos(f*x + e)^2*log(-sin(f*x + e) + 1) + 2*((a^2*b^2 - b^4)*d^3 + 2*(3*(a^2*b^2 -
b^4)*c*d^2 - (a^3*b - a*b^3)*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*b^3 - b^5)*f*cos(f*x + e)^2)]

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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 )^{3} \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))**3/(a+b*sec(f*x+e)),x)

[Out]

Integral((c + d*sec(e + f*x))**3*sec(e + f*x)/(a + b*sec(e + f*x)), x)

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Giac [B]  time = 1.34235, size = 474, normalized size = 2.79 \begin{align*} \frac{\frac{{\left (6 \, b^{2} c^{2} d - 6 \, a b c d^{2} + 2 \, a^{2} d^{3} + b^{2} d^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{b^{3}} - \frac{{\left (6 \, b^{2} c^{2} d - 6 \, a b c d^{2} + 2 \, a^{2} d^{3} + b^{2} d^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{b^{3}} - \frac{4 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\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^{3}} - \frac{2 \,{\left (6 \, b c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - b d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 6 \, b c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, a d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - b d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{2} b^{2}}}{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))^3/(a+b*sec(f*x+e)),x, algorithm="giac")

[Out]

1/2*((6*b^2*c^2*d - 6*a*b*c*d^2 + 2*a^2*d^3 + b^2*d^3)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/b^3 - (6*b^2*c^2*d -
6*a*b*c*d^2 + 2*a^2*d^3 + b^2*d^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^3 - 4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^
2*b*c*d^2 - a^3*d^3)*(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^3) - 2*(6*b*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 2*a*d^3*t
an(1/2*f*x + 1/2*e)^3 - b*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*b*c*d^2*tan(1/2*f*x + 1/2*e) + 2*a*d^3*tan(1/2*f*x +
1/2*e) - b*d^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*b^2))/f