3.260 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^3}{(a+b \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=228 \[ -\frac{(b c-a d)^3 \sin (e+f x)}{b^2 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac{d^2 (3 b c-2 a d) \tanh ^{-1}(\sin (e+f x))}{b^3 f}+\frac{2 (b c-a d)^2 (2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^3 f \sqrt{a-b} \sqrt{a+b}}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b f (a-b)^{3/2} (a+b)^{3/2}}+\frac{d^3 \tan (e+f x)}{b^2 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.467264, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.29, Rules used = {3988, 2952, 2664, 12, 2659, 208, 3770, 3767, 8} \[ -\frac{(b c-a d)^3 \sin (e+f x)}{b^2 f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac{d^2 (3 b c-2 a d) \tanh ^{-1}(\sin (e+f x))}{b^3 f}+\frac{2 (b c-a d)^2 (2 a d+b c) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^3 f \sqrt{a-b} \sqrt{a+b}}+\frac{2 (b c-a d)^3 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b f (a-b)^{3/2} (a+b)^{3/2}}+\frac{d^3 \tan (e+f x)}{b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(3*b*c - 2*a*d)*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]])/(a*(a - b)^(3/2)*b*(a + b)^(3/2)*f) + (2*(b*c - a*d)^2*(b*c + 2*a*d)*ArcTanh[(Sqrt[a - b]*Tan[(e
+ f*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*b^3*Sqrt[a + b]*f) - ((b*c - a*d)^3*Sin[e + f*x])/(b^2*(a^2 - b^2)*f*
(b + a*Cos[e + f*x])) + (d^3*Tan[e + f*x])/(b^2*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 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 1.57867, size = 362, normalized size = 1.59 \[ \frac{\cos (e+f x) (a \cos (e+f x)+b) (c+d \sec (e+f x))^3 \left (-\frac{2 (b c-a d)^2 \left (2 a^2 d+a b c-3 b^2 d\right ) (a \cos (e+f x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+d^2 (2 a d-3 b c) (a \cos (e+f x)+b) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+d^2 (3 b c-2 a d) (a \cos (e+f x)+b) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+\frac{b (b c-a d)^3 \sin (e+f x)}{(b-a) (a+b)}+\frac{b d^3 \sin \left (\frac{1}{2} (e+f x)\right ) (a \cos (e+f x)+b)}{\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )}+\frac{b d^3 \sin \left (\frac{1}{2} (e+f x)\right ) (a \cos (e+f x)+b)}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}\right )}{b^3 f (a+b \sec (e+f x))^2 (c \cos (e+f x)+d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.092, size = 790, 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))^2,x)

[Out]

-2/f/b^2/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a^3*d^3+6/f/b/(a^2-b
^2)*tan(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a^2*c*d^2-6/f/(a^2-b^2)*tan(1/2*f*x
+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a*c^2*d+2/f*b/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(tan(1/
2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*c^3+4/f/b^3/(a+b)/(a-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^4*d^3-6/f/b^2/(a+b)/(a-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^3*c*d^2-6/f/b/(a+b)/(a-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^2*d^3+2/f/(a+b)/(a-b)/((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*a+12/f/(a+b)/(a-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-
6/f*b/(a+b)/(a-b)/((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*d-1/f*d^3/b^2/
(tan(1/2*f*x+1/2*e)+1)-2/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)+1)*a+3/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)+1)*c-1/f*d^3/b
^2/(tan(1/2*f*x+1/2*e)-1)+2/f*d^3/b^3*ln(tan(1/2*f*x+1/2*e)-1)*a-3/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)-1)*c

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 154.86, size = 2800, normalized size = 12.28 \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))^2,x, algorithm="fricas")

[Out]

[1/2*(((a^2*b^3*c^3 - 3*a*b^4*c^2*d - 3*(a^4*b - 2*a^2*b^3)*c*d^2 + (2*a^5 - 3*a^3*b^2)*d^3)*cos(f*x + e)^2 +
(a*b^4*c^3 - 3*b^5*c^2*d - 3*(a^3*b^2 - 2*a*b^4)*c*d^2 + (2*a^4*b - 3*a^2*b^3)*d^3)*cos(f*x + e))*sqrt(a^2 - b
^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)) + ((3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2
*(a^6 - 2*a^4*b^2 + a^2*b^4)*d^3)*cos(f*x + e)^2 + (3*(a^4*b^2 - 2*a^2*b^4 + b^6)*c*d^2 - 2*(a^5*b - 2*a^3*b^3
+ a*b^5)*d^3)*cos(f*x + e))*log(sin(f*x + e) + 1) - ((3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2*(a^6 - 2*a^4*b^
2 + a^2*b^4)*d^3)*cos(f*x + e)^2 + (3*(a^4*b^2 - 2*a^2*b^4 + b^6)*c*d^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d^3)*c
os(f*x + e))*log(-sin(f*x + e) + 1) + 2*((a^4*b^2 - 2*a^2*b^4 + b^6)*d^3 - ((a^2*b^4 - b^6)*c^3 - 3*(a^3*b^3 -
a*b^5)*c^2*d + 3*(a^4*b^2 - a^2*b^4)*c*d^2 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*d^3)*cos(f*x + e))*sin(f*x + e))/(
(a^5*b^3 - 2*a^3*b^5 + a*b^7)*f*cos(f*x + e)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*f*cos(f*x + e)), 1/2*(2*((a^2*b^3
*c^3 - 3*a*b^4*c^2*d - 3*(a^4*b - 2*a^2*b^3)*c*d^2 + (2*a^5 - 3*a^3*b^2)*d^3)*cos(f*x + e)^2 + (a*b^4*c^3 - 3*
b^5*c^2*d - 3*(a^3*b^2 - 2*a*b^4)*c*d^2 + (2*a^4*b - 3*a^2*b^3)*d^3)*cos(f*x + e))*sqrt(-a^2 + b^2)*arctan(-sq
rt(-a^2 + b^2)*(b*cos(f*x + e) + a)/((a^2 - b^2)*sin(f*x + e))) + ((3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2*(a
^6 - 2*a^4*b^2 + a^2*b^4)*d^3)*cos(f*x + e)^2 + (3*(a^4*b^2 - 2*a^2*b^4 + b^6)*c*d^2 - 2*(a^5*b - 2*a^3*b^3 +
a*b^5)*d^3)*cos(f*x + e))*log(sin(f*x + e) + 1) - ((3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2*(a^6 - 2*a^4*b^2 +
a^2*b^4)*d^3)*cos(f*x + e)^2 + (3*(a^4*b^2 - 2*a^2*b^4 + b^6)*c*d^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d^3)*cos(
f*x + e))*log(-sin(f*x + e) + 1) + 2*((a^4*b^2 - 2*a^2*b^4 + b^6)*d^3 - ((a^2*b^4 - b^6)*c^3 - 3*(a^3*b^3 - a*
b^5)*c^2*d + 3*(a^4*b^2 - a^2*b^4)*c*d^2 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*d^3)*cos(f*x + e))*sin(f*x + e))/((a^
5*b^3 - 2*a^3*b^5 + a*b^7)*f*cos(f*x + e)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*f*cos(f*x + e))]

________________________________________________________________________________________

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 )}}{\left (a + b \sec{\left (e + f x \right )}\right )^{2}}\, 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))**2,x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.43682, size = 757, normalized size = 3.32 \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))^2,x, algorithm="giac")

[Out]

-(2*(a*b^3*c^3 - 3*b^4*c^2*d - 3*a^3*b*c*d^2 + 6*a*b^3*c*d^2 + 2*a^4*d^3 - 3*a^2*b^2*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)))/((a
^2*b^3 - b^5)*sqrt(-a^2 + b^2)) - 2*(b^3*c^3*tan(1/2*f*x + 1/2*e)^3 - 3*a*b^2*c^2*d*tan(1/2*f*x + 1/2*e)^3 + 3
*a^2*b*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 2*a^3*d^3*tan(1/2*f*x + 1/2*e)^3 + a^2*b*d^3*tan(1/2*f*x + 1/2*e)^3 + a*
b^2*d^3*tan(1/2*f*x + 1/2*e)^3 - b^3*d^3*tan(1/2*f*x + 1/2*e)^3 - b^3*c^3*tan(1/2*f*x + 1/2*e) + 3*a*b^2*c^2*d
*tan(1/2*f*x + 1/2*e) - 3*a^2*b*c*d^2*tan(1/2*f*x + 1/2*e) + 2*a^3*d^3*tan(1/2*f*x + 1/2*e) + a^2*b*d^3*tan(1/
2*f*x + 1/2*e) - a*b^2*d^3*tan(1/2*f*x + 1/2*e) - b^3*d^3*tan(1/2*f*x + 1/2*e))/((a*tan(1/2*f*x + 1/2*e)^4 - b
*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + a + b)*(a^2*b^2 - b^4)) - (3*b*c*d^2 - 2*a*d^3)*log(abs
(tan(1/2*f*x + 1/2*e) + 1))/b^3 + (3*b*c*d^2 - 2*a*d^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^3)/f