3.252 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^4}{a+b \sec (e+f x)} \, dx\)

Optimal. Leaf size=247 \[ \frac{d^2 \left (a^2 d^2-4 a b c d+6 b^2 c^2\right ) \tan (e+f x)}{b^3 f}+\frac{d (2 b c-a d) \left (a^2 d^2-2 a b c d+2 b^2 c^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac{d^3 (4 b c-a d) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac{d^3 (4 b c-a d) \tan (e+f x) \sec (e+f x)}{2 b^2 f}+\frac{2 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b^4 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^4 \tan ^3(e+f x)}{3 b f}+\frac{d^4 \tan (e+f x)}{b f} \]

[Out]

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

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Rubi [A]  time = 0.444554, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ \frac{d^2 \left (a^2 d^2-4 a b c d+6 b^2 c^2\right ) \tan (e+f x)}{b^3 f}+\frac{d (2 b c-a d) \left (a^2 d^2-2 a b c d+2 b^2 c^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac{d^3 (4 b c-a d) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac{d^3 (4 b c-a d) \tan (e+f x) \sec (e+f x)}{2 b^2 f}+\frac{2 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{b^4 f \sqrt{a-b} \sqrt{a+b}}+\frac{d^4 \tan ^3(e+f x)}{3 b f}+\frac{d^4 \tan (e+f x)}{b f} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + b*Sec[e + f*x]),x]

[Out]

(d^3*(4*b*c - a*d)*ArcTanh[Sin[e + f*x]])/(2*b^2*f) + (d*(2*b*c - a*d)*(2*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*ArcTa
nh[Sin[e + f*x]])/(b^4*f) + (2*(b*c - a*d)^4*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]
*b^4*Sqrt[a + b]*f) + (d^4*Tan[e + f*x])/(b*f) + (d^2*(6*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Tan[e + f*x])/(b^3*f)
+ (d^3*(4*b*c - a*d)*Sec[e + f*x]*Tan[e + f*x])/(2*b^2*f) + (d^4*Tan[e + f*x]^3)/(3*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))^4}{a+b \sec (e+f x)} \, dx &=\int \frac{(d+c \cos (e+f x))^4 \sec ^4(e+f x)}{b+a \cos (e+f x)} \, dx\\ &=\int \left (\frac{(b c-a d)^4}{b^4 (b+a \cos (e+f x))}+\frac{d (2 b c-a d) \left (2 b^2 c^2-2 a b c d+a^2 d^2\right ) \sec (e+f x)}{b^4}+\frac{d^2 \left (6 b^2 c^2-4 a b c d+a^2 d^2\right ) \sec ^2(e+f x)}{b^3}+\frac{d^3 (4 b c-a d) \sec ^3(e+f x)}{b^2}+\frac{d^4 \sec ^4(e+f x)}{b}\right ) \, dx\\ &=\frac{d^4 \int \sec ^4(e+f x) \, dx}{b}+\frac{(b c-a d)^4 \int \frac{1}{b+a \cos (e+f x)} \, dx}{b^4}+\frac{\left (d^3 (4 b c-a d)\right ) \int \sec ^3(e+f x) \, dx}{b^2}+\frac{\left (d^2 \left (6 b^2 c^2-4 a b c d+a^2 d^2\right )\right ) \int \sec ^2(e+f x) \, dx}{b^3}+\frac{\left (d (2 b c-a d) \left (2 b^2 c^2-2 a b c d+a^2 d^2\right )\right ) \int \sec (e+f x) \, dx}{b^4}\\ &=\frac{d (2 b c-a d) \left (2 b^2 c^2-2 a b c d+a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac{d^3 (4 b c-a d) \sec (e+f x) \tan (e+f x)}{2 b^2 f}+\frac{\left (d^3 (4 b c-a d)\right ) \int \sec (e+f x) \, dx}{2 b^2}-\frac{d^4 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{b f}+\frac{\left (2 (b c-a d)^4\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^4 f}-\frac{\left (d^2 \left (6 b^2 c^2-4 a b c d+a^2 d^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{b^3 f}\\ &=\frac{d^3 (4 b c-a d) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac{d (2 b c-a d) \left (2 b^2 c^2-2 a b c d+a^2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{b^4 f}+\frac{2 (b c-a d)^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^4 \sqrt{a+b} f}+\frac{d^4 \tan (e+f x)}{b f}+\frac{d^2 \left (6 b^2 c^2-4 a b c d+a^2 d^2\right ) \tan (e+f x)}{b^3 f}+\frac{d^3 (4 b c-a d) \sec (e+f x) \tan (e+f x)}{2 b^2 f}+\frac{d^4 \tan ^3(e+f x)}{3 b f}\\ \end{align*}

Mathematica [B]  time = 4.34563, size = 580, normalized size = 2.35 \[ \frac{\cos ^3(e+f x) (a \cos (e+f x)+b) (c+d \sec (e+f x))^4 \left (\frac{4 b d^2 \left (3 a^2 d^2-12 a b c d+2 b^2 \left (9 c^2+d^2\right )\right ) \sin \left (\frac{1}{2} (e+f x)\right )}{\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )}+\frac{4 b d^2 \left (3 a^2 d^2-12 a b c d+2 b^2 \left (9 c^2+d^2\right )\right ) \sin \left (\frac{1}{2} (e+f x)\right )}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}-6 d \left (8 a^2 b c d^2-2 a^3 d^3-a b^2 d \left (12 c^2+d^2\right )+4 b^3 c \left (2 c^2+d^2\right )\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-6 d \left (-8 a^2 b c d^2+2 a^3 d^3+a b^2 d \left (12 c^2+d^2\right )-4 b^3 c \left (2 c^2+d^2\right )\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-\frac{24 (b c-a d)^4 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{b^2 d^3 (b (12 c+d)-3 a d)}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{b^2 d^3 (b (12 c+d)-3 a d)}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}+\frac{2 b^3 d^4 \sin \left (\frac{1}{2} (e+f x)\right )}{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3}+\frac{2 b^3 d^4 \sin \left (\frac{1}{2} (e+f x)\right )}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}\right )}{12 b^4 f (a+b \sec (e+f x)) (c \cos (e+f x)+d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^4)/(a + b*Sec[e + f*x]),x]

[Out]

(Cos[e + f*x]^3*(b + a*Cos[e + f*x])*(c + d*Sec[e + f*x])^4*((-24*(b*c - a*d)^4*ArcTanh[((-a + b)*Tan[(e + f*x
)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - 6*d*(8*a^2*b*c*d^2 - 2*a^3*d^3 + 4*b^3*c*(2*c^2 + d^2) - a*b^2*d*(12
*c^2 + d^2))*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - 6*d*(-8*a^2*b*c*d^2 + 2*a^3*d^3 - 4*b^3*c*(2*c^2 + d^2
) + a*b^2*d*(12*c^2 + d^2))*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (b^2*d^3*(-3*a*d + b*(12*c + d)))/(Cos[
(e + f*x)/2] - Sin[(e + f*x)/2])^2 + (2*b^3*d^4*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + (4
*b*d^2*(-12*a*b*c*d + 3*a^2*d^2 + 2*b^2*(9*c^2 + d^2))*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])
+ (2*b^3*d^4*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - (b^2*d^3*(-3*a*d + b*(12*c + d)))/(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (4*b*d^2*(-12*a*b*c*d + 3*a^2*d^2 + 2*b^2*(9*c^2 + d^2))*Sin[(e + f*x)
/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(12*b^4*f*(d + c*Cos[e + f*x])^4*(a + b*Sec[e + f*x]))

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Maple [B]  time = 0.098, size = 1066, normalized size = 4.3 \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))^4/(a+b*sec(f*x+e)),x)

[Out]

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^4-1/3/f*d^4/b/(tan(1/2*f*x+1/2
*e)+1)^3-1/f*d^4/b/(tan(1/2*f*x+1/2*e)+1)+1/2/f*d^4/b/(tan(1/2*f*x+1/2*e)+1)^2-1/3/f*d^4/b/(tan(1/2*f*x+1/2*e)
-1)^3-1/f*d^4/b/(tan(1/2*f*x+1/2*e)-1)-1/2/f*d^4/b/(tan(1/2*f*x+1/2*e)-1)^2+12/f/b^2/((a+b)*(a-b))^(1/2)*arcta
nh((a-b)*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a^2*c^2*d^2-8/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*c*d^3-8/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^3*d+2/f*d^3/b/(tan(1/2*f*x+1/2*e)-1)^2*c-1/2/f*d^4/b^2/(tan(1/2*f*x+1/2*e)+1)*a-6/f*d^2/b/(ta
n(1/2*f*x+1/2*e)+1)*c^2+2/f*d^3/b/(tan(1/2*f*x+1/2*e)+1)*c+1/2/f*d^4/b^2/(tan(1/2*f*x+1/2*e)+1)^2*a-2/f*d^3/b/
(tan(1/2*f*x+1/2*e)+1)^2*c-1/f*d^4/b^4*ln(tan(1/2*f*x+1/2*e)+1)*a^3-1/2/f*d^4/b^2*ln(tan(1/2*f*x+1/2*e)+1)*a+4
/f*d/b*ln(tan(1/2*f*x+1/2*e)+1)*c^3+2/f*d^3/b*ln(tan(1/2*f*x+1/2*e)+1)*c-1/f*d^4/b^3/(tan(1/2*f*x+1/2*e)+1)*a^
2+1/f*d^4/b^4*ln(tan(1/2*f*x+1/2*e)-1)*a^3+1/2/f*d^4/b^2*ln(tan(1/2*f*x+1/2*e)-1)*a-4/f*d/b*ln(tan(1/2*f*x+1/2
*e)-1)*c^3-2/f*d^3/b*ln(tan(1/2*f*x+1/2*e)-1)*c-1/f*d^4/b^3/(tan(1/2*f*x+1/2*e)-1)*a^2-1/2/f*d^4/b^2/(tan(1/2*
f*x+1/2*e)-1)*a-6/f*d^2/b/(tan(1/2*f*x+1/2*e)-1)*c^2+2/f*d^3/b/(tan(1/2*f*x+1/2*e)-1)*c-1/2/f*d^4/b^2/(tan(1/2
*f*x+1/2*e)-1)^2*a+6/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)-1)*a*c^2+4/f*d^3/b^2/(tan(1/2*f*x+1/2*e)-1)*a*c+2/f/b^4/(
(a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a^4*d^4+4/f*d^3/b^3*ln(tan(1/2*f*x+1/
2*e)+1)*a^2*c-6/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)+1)*a*c^2+4/f*d^3/b^2/(tan(1/2*f*x+1/2*e)+1)*a*c-4/f*d^3/b^3*ln
(tan(1/2*f*x+1/2*e)-1)*a^2*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))^4/(a+b*sec(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^4/(a+b*sec(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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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 )^{4} \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))**4/(a+b*sec(f*x+e)),x)

[Out]

Integral((c + d*sec(e + f*x))**4*sec(e + f*x)/(a + b*sec(e + f*x)), x)

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Giac [B]  time = 1.46134, size = 848, normalized size = 3.43 \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))^4/(a+b*sec(f*x+e)),x, algorithm="giac")

[Out]

1/6*(3*(8*b^3*c^3*d - 12*a*b^2*c^2*d^2 + 8*a^2*b*c*d^3 + 4*b^3*c*d^3 - 2*a^3*d^4 - a*b^2*d^4)*log(abs(tan(1/2*
f*x + 1/2*e) + 1))/b^4 - 3*(8*b^3*c^3*d - 12*a*b^2*c^2*d^2 + 8*a^2*b*c*d^3 + 4*b^3*c*d^3 - 2*a^3*d^4 - a*b^2*d
^4)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^4 - 12*(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 +
a^4*d^4)*(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^4) - 2*(36*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^5 - 24*a*b*c*d^3*ta
n(1/2*f*x + 1/2*e)^5 - 12*b^2*c*d^3*tan(1/2*f*x + 1/2*e)^5 + 6*a^2*d^4*tan(1/2*f*x + 1/2*e)^5 + 3*a*b*d^4*tan(
1/2*f*x + 1/2*e)^5 + 6*b^2*d^4*tan(1/2*f*x + 1/2*e)^5 - 72*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e)^3 + 48*a*b*c*d^3*t
an(1/2*f*x + 1/2*e)^3 - 12*a^2*d^4*tan(1/2*f*x + 1/2*e)^3 - 4*b^2*d^4*tan(1/2*f*x + 1/2*e)^3 + 36*b^2*c^2*d^2*
tan(1/2*f*x + 1/2*e) - 24*a*b*c*d^3*tan(1/2*f*x + 1/2*e) + 12*b^2*c*d^3*tan(1/2*f*x + 1/2*e) + 6*a^2*d^4*tan(1
/2*f*x + 1/2*e) - 3*a*b*d^4*tan(1/2*f*x + 1/2*e) + 6*b^2*d^4*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 -
1)^3*b^3))/f