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

Optimal. Leaf size=237 \[ \frac{\left (2 a c^3+3 a c d^2-4 b c^2 d-b d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f (c-d)^{7/2} (c+d)^{7/2}}+\frac{\left (-11 a c^2 d-4 a d^3+2 b c^3+13 b c d^2\right ) \tan (e+f x)}{6 f \left (c^2-d^2\right )^3 (c+d \sec (e+f x))}+\frac{\left (-5 a c d+2 b c^2+3 b d^2\right ) \tan (e+f x)}{6 f \left (c^2-d^2\right )^2 (c+d \sec (e+f x))^2}+\frac{(b c-a d) \tan (e+f x)}{3 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^3} \]

[Out]

((2*a*c^3 - 4*b*c^2*d + 3*a*c*d^2 - b*d^3)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/((c - d)^(7/2)
*(c + d)^(7/2)*f) + ((b*c - a*d)*Tan[e + f*x])/(3*(c^2 - d^2)*f*(c + d*Sec[e + f*x])^3) + ((2*b*c^2 - 5*a*c*d
+ 3*b*d^2)*Tan[e + f*x])/(6*(c^2 - d^2)^2*f*(c + d*Sec[e + f*x])^2) + ((2*b*c^3 - 11*a*c^2*d + 13*b*c*d^2 - 4*
a*d^3)*Tan[e + f*x])/(6*(c^2 - d^2)^3*f*(c + d*Sec[e + f*x]))

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Rubi [A]  time = 0.511888, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {4003, 12, 3831, 2659, 208} \[ \frac{\left (2 a c^3+3 a c d^2-4 b c^2 d-b d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{f (c-d)^{7/2} (c+d)^{7/2}}+\frac{\left (-11 a c^2 d-4 a d^3+2 b c^3+13 b c d^2\right ) \tan (e+f x)}{6 f \left (c^2-d^2\right )^3 (c+d \sec (e+f x))}+\frac{\left (-5 a c d+2 b c^2+3 b d^2\right ) \tan (e+f x)}{6 f \left (c^2-d^2\right )^2 (c+d \sec (e+f x))^2}+\frac{(b c-a d) \tan (e+f x)}{3 f \left (c^2-d^2\right ) (c+d \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*c^3 - 4*b*c^2*d + 3*a*c*d^2 - b*d^3)*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/((c - d)^(7/2)
*(c + d)^(7/2)*f) + ((b*c - a*d)*Tan[e + f*x])/(3*(c^2 - d^2)*f*(c + d*Sec[e + f*x])^3) + ((2*b*c^2 - 5*a*c*d
+ 3*b*d^2)*Tan[e + f*x])/(6*(c^2 - d^2)^2*f*(c + d*Sec[e + f*x])^2) + ((2*b*c^3 - 11*a*c^2*d + 13*b*c*d^2 - 4*
a*d^3)*Tan[e + f*x])/(6*(c^2 - d^2)^3*f*(c + d*Sec[e + f*x]))

Rule 4003

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a*B
)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] &
& LtQ[m, -1]

Rule 12

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

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]

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (a+b \sec (e+f x))}{(c+d \sec (e+f x))^4} \, dx &=\frac{(b c-a d) \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}-\frac{\int \frac{\sec (e+f x) (-3 (a c-b d)-2 (b c-a d) \sec (e+f x))}{(c+d \sec (e+f x))^3} \, dx}{3 \left (c^2-d^2\right )}\\ &=\frac{(b c-a d) \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{\left (2 b c^2-5 a c d+3 b d^2\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))^2}+\frac{\int \frac{\sec (e+f x) \left (2 \left (3 a c^2-5 b c d+2 a d^2\right )+\left (2 b c^2-5 a c d+3 b d^2\right ) \sec (e+f x)\right )}{(c+d \sec (e+f x))^2} \, dx}{6 \left (c^2-d^2\right )^2}\\ &=\frac{(b c-a d) \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{\left (2 b c^2-5 a c d+3 b d^2\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))^2}+\frac{\left (2 b c^3-11 a c^2 d+13 b c d^2-4 a d^3\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^3 f (c+d \sec (e+f x))}-\frac{\int -\frac{3 \left (2 a c^3-4 b c^2 d+3 a c d^2-b d^3\right ) \sec (e+f x)}{c+d \sec (e+f x)} \, dx}{6 \left (c^2-d^2\right )^3}\\ &=\frac{(b c-a d) \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{\left (2 b c^2-5 a c d+3 b d^2\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))^2}+\frac{\left (2 b c^3-11 a c^2 d+13 b c d^2-4 a d^3\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^3 f (c+d \sec (e+f x))}+\frac{\left (2 a c^3-4 b c^2 d+3 a c d^2-b d^3\right ) \int \frac{\sec (e+f x)}{c+d \sec (e+f x)} \, dx}{2 \left (c^2-d^2\right )^3}\\ &=\frac{(b c-a d) \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{\left (2 b c^2-5 a c d+3 b d^2\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))^2}+\frac{\left (2 b c^3-11 a c^2 d+13 b c d^2-4 a d^3\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^3 f (c+d \sec (e+f x))}+\frac{\left (2 a c^3-4 b c^2 d+3 a c d^2-b d^3\right ) \int \frac{1}{1+\frac{c \cos (e+f x)}{d}} \, dx}{2 d \left (c^2-d^2\right )^3}\\ &=\frac{(b c-a d) \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{\left (2 b c^2-5 a c d+3 b d^2\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))^2}+\frac{\left (2 b c^3-11 a c^2 d+13 b c d^2-4 a d^3\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^3 f (c+d \sec (e+f x))}+\frac{\left (2 a c^3-4 b c^2 d+3 a c d^2-b d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{c}{d}+\left (1-\frac{c}{d}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{d \left (c^2-d^2\right )^3 f}\\ &=\frac{\left (2 a c^3-4 b c^2 d+3 a c d^2-b d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{(c-d)^{7/2} (c+d)^{7/2} f}+\frac{(b c-a d) \tan (e+f x)}{3 \left (c^2-d^2\right ) f (c+d \sec (e+f x))^3}+\frac{\left (2 b c^2-5 a c d+3 b d^2\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^2 f (c+d \sec (e+f x))^2}+\frac{\left (2 b c^3-11 a c^2 d+13 b c d^2-4 a d^3\right ) \tan (e+f x)}{6 \left (c^2-d^2\right )^3 f (c+d \sec (e+f x))}\\ \end{align*}

Mathematica [A]  time = 1.03578, size = 405, normalized size = 1.71 \[ \frac{\sec ^3(e+f x) (a+b \sec (e+f x)) (c \cos (e+f x)+d) \left (\frac{24 \left (a \left (2 c^3+3 c d^2\right )-b d \left (4 c^2+d^2\right )\right ) (c \cos (e+f x)+d)^3 \tanh ^{-1}\left (\frac{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}+54 a c^3 d^2 \sin (2 (e+f x))+39 a c^2 d^3 \sin (e+f x)-5 a c^2 d^3 \sin (3 (e+f x))+18 a c^4 d \sin (e+f x)+18 a c^4 d \sin (3 (e+f x))+6 a c d^4 \sin (2 (e+f x))+18 a d^5 \sin (e+f x)+2 a d^5 \sin (3 (e+f x))-18 b c^3 d^2 \sin (e+f x)-10 b c^3 d^2 \sin (3 (e+f x))-54 b c^2 d^3 \sin (2 (e+f x))-12 b c^4 d \sin (2 (e+f x))-6 b c^5 \sin (e+f x)-6 b c^5 \sin (3 (e+f x))-51 b c d^4 \sin (e+f x)+b c d^4 \sin (3 (e+f x))+6 b d^5 \sin (2 (e+f x))\right )}{24 f \left (d^2-c^2\right )^3 (a \cos (e+f x)+b) (c+d \sec (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + c*Cos[e + f*x])*Sec[e + f*x]^3*(a + b*Sec[e + f*x])*((24*(-(b*d*(4*c^2 + d^2)) + a*(2*c^3 + 3*c*d^2))*Ar
cTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(d + c*Cos[e + f*x])^3)/Sqrt[c^2 - d^2] - 6*b*c^5*Sin[e + f
*x] + 18*a*c^4*d*Sin[e + f*x] - 18*b*c^3*d^2*Sin[e + f*x] + 39*a*c^2*d^3*Sin[e + f*x] - 51*b*c*d^4*Sin[e + f*x
] + 18*a*d^5*Sin[e + f*x] - 12*b*c^4*d*Sin[2*(e + f*x)] + 54*a*c^3*d^2*Sin[2*(e + f*x)] - 54*b*c^2*d^3*Sin[2*(
e + f*x)] + 6*a*c*d^4*Sin[2*(e + f*x)] + 6*b*d^5*Sin[2*(e + f*x)] - 6*b*c^5*Sin[3*(e + f*x)] + 18*a*c^4*d*Sin[
3*(e + f*x)] - 10*b*c^3*d^2*Sin[3*(e + f*x)] - 5*a*c^2*d^3*Sin[3*(e + f*x)] + b*c*d^4*Sin[3*(e + f*x)] + 2*a*d
^5*Sin[3*(e + f*x)]))/(24*(-c^2 + d^2)^3*f*(b + a*Cos[e + f*x])*(c + d*Sec[e + f*x])^4)

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Maple [A]  time = 0.092, size = 376, normalized size = 1.6 \begin{align*}{\frac{1}{f} \left ( -2\,{\frac{1}{ \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}c- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}d-c-d \right ) ^{3}} \left ( -1/2\,{\frac{ \left ( 6\,a{c}^{2}d+3\,ac{d}^{2}+2\,a{d}^{3}-2\,b{c}^{3}-2\,b{c}^{2}d-6\,bc{d}^{2}-b{d}^{3} \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{5}}{ \left ( c-d \right ) \left ({c}^{3}+3\,{c}^{2}d+3\,{d}^{2}c+{d}^{3} \right ) }}+2/3\,{\frac{ \left ( 9\,a{c}^{2}d+a{d}^{3}-3\,b{c}^{3}-7\,bc{d}^{2} \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{3}}{ \left ({c}^{2}+2\,cd+{d}^{2} \right ) \left ({c}^{2}-2\,cd+{d}^{2} \right ) }}-1/2\,{\frac{ \left ( 6\,a{c}^{2}d-3\,ac{d}^{2}+2\,a{d}^{3}-2\,b{c}^{3}+2\,b{c}^{2}d-6\,bc{d}^{2}+b{d}^{3} \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{ \left ( c+d \right ) \left ({c}^{3}-3\,{c}^{2}d+3\,{d}^{2}c-{d}^{3} \right ) }} \right ) }+{\frac{2\,a{c}^{3}+3\,ac{d}^{2}-4\,b{c}^{2}d-b{d}^{3}}{{c}^{6}-3\,{c}^{4}{d}^{2}+3\,{c}^{2}{d}^{4}-{d}^{6}}{\it Artanh} \left ({(c-d)\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ){\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}}} \right ){\frac{1}{\sqrt{ \left ( c+d \right ) \left ( c-d \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-2*(-1/2*(6*a*c^2*d+3*a*c*d^2+2*a*d^3-2*b*c^3-2*b*c^2*d-6*b*c*d^2-b*d^3)/(c-d)/(c^3+3*c^2*d+3*c*d^2+d^3)*
tan(1/2*f*x+1/2*e)^5+2/3*(9*a*c^2*d+a*d^3-3*b*c^3-7*b*c*d^2)/(c^2+2*c*d+d^2)/(c^2-2*c*d+d^2)*tan(1/2*f*x+1/2*e
)^3-1/2*(6*a*c^2*d-3*a*c*d^2+2*a*d^3-2*b*c^3+2*b*c^2*d-6*b*c*d^2+b*d^3)/(c+d)/(c^3-3*c^2*d+3*c*d^2-d^3)*tan(1/
2*f*x+1/2*e))/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^3+(2*a*c^3+3*a*c*d^2-4*b*c^2*d-b*d^3)/(c^6-3
*c^4*d^2+3*c^2*d^4-d^6)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^(1/2)))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.751108, size = 2707, normalized size = 11.42 \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)*(a+b*sec(f*x+e))/(c+d*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

[1/12*(3*(2*a*c^3*d^3 - 4*b*c^2*d^4 + 3*a*c*d^5 - b*d^6 + (2*a*c^6 - 4*b*c^5*d + 3*a*c^4*d^2 - b*c^3*d^3)*cos(
f*x + e)^3 + 3*(2*a*c^5*d - 4*b*c^4*d^2 + 3*a*c^3*d^3 - b*c^2*d^4)*cos(f*x + e)^2 + 3*(2*a*c^4*d^2 - 4*b*c^3*d
^3 + 3*a*c^2*d^4 - b*c*d^5)*cos(f*x + e))*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)
^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f*x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x +
e) + d^2)) + 2*(2*b*c^5*d^2 - 11*a*c^4*d^3 + 11*b*c^3*d^4 + 7*a*c^2*d^5 - 13*b*c*d^6 + 4*a*d^7 + (6*b*c^7 - 18
*a*c^6*d + 4*b*c^5*d^2 + 23*a*c^4*d^3 - 11*b*c^3*d^4 - 7*a*c^2*d^5 + b*c*d^6 + 2*a*d^7)*cos(f*x + e)^2 + 3*(2*
b*c^6*d - 9*a*c^5*d^2 + 7*b*c^4*d^3 + 8*a*c^3*d^4 - 10*b*c^2*d^5 + a*c*d^6 + b*d^7)*cos(f*x + e))*sin(f*x + e)
)/((c^11 - 4*c^9*d^2 + 6*c^7*d^4 - 4*c^5*d^6 + c^3*d^8)*f*cos(f*x + e)^3 + 3*(c^10*d - 4*c^8*d^3 + 6*c^6*d^5 -
4*c^4*d^7 + c^2*d^9)*f*cos(f*x + e)^2 + 3*(c^9*d^2 - 4*c^7*d^4 + 6*c^5*d^6 - 4*c^3*d^8 + c*d^10)*f*cos(f*x +
e) + (c^8*d^3 - 4*c^6*d^5 + 6*c^4*d^7 - 4*c^2*d^9 + d^11)*f), 1/6*(3*(2*a*c^3*d^3 - 4*b*c^2*d^4 + 3*a*c*d^5 -
b*d^6 + (2*a*c^6 - 4*b*c^5*d + 3*a*c^4*d^2 - b*c^3*d^3)*cos(f*x + e)^3 + 3*(2*a*c^5*d - 4*b*c^4*d^2 + 3*a*c^3*
d^3 - b*c^2*d^4)*cos(f*x + e)^2 + 3*(2*a*c^4*d^2 - 4*b*c^3*d^3 + 3*a*c^2*d^4 - b*c*d^5)*cos(f*x + e))*sqrt(-c^
2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*sin(f*x + e))) + (2*b*c^5*d^2 - 11*a*c^4*d
^3 + 11*b*c^3*d^4 + 7*a*c^2*d^5 - 13*b*c*d^6 + 4*a*d^7 + (6*b*c^7 - 18*a*c^6*d + 4*b*c^5*d^2 + 23*a*c^4*d^3 -
11*b*c^3*d^4 - 7*a*c^2*d^5 + b*c*d^6 + 2*a*d^7)*cos(f*x + e)^2 + 3*(2*b*c^6*d - 9*a*c^5*d^2 + 7*b*c^4*d^3 + 8*
a*c^3*d^4 - 10*b*c^2*d^5 + a*c*d^6 + b*d^7)*cos(f*x + e))*sin(f*x + e))/((c^11 - 4*c^9*d^2 + 6*c^7*d^4 - 4*c^5
*d^6 + c^3*d^8)*f*cos(f*x + e)^3 + 3*(c^10*d - 4*c^8*d^3 + 6*c^6*d^5 - 4*c^4*d^7 + c^2*d^9)*f*cos(f*x + e)^2 +
3*(c^9*d^2 - 4*c^7*d^4 + 6*c^5*d^6 - 4*c^3*d^8 + c*d^10)*f*cos(f*x + e) + (c^8*d^3 - 4*c^6*d^5 + 6*c^4*d^7 -
4*c^2*d^9 + d^11)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sec{\left (e + f x \right )}\right ) \sec{\left (e + f x \right )}}{\left (c + d \sec{\left (e + f x \right )}\right )^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.4312, size = 980, normalized size = 4.14 \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)*(a+b*sec(f*x+e))/(c+d*sec(f*x+e))^4,x, algorithm="giac")

[Out]

-1/3*(3*(2*a*c^3 - 4*b*c^2*d + 3*a*c*d^2 - b*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c
*tan(1/2*f*x + 1/2*e) - d*tan(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))/((c^6 - 3*c^4*d^2 + 3*c^2*d^4 - d^6)*sqrt(-
c^2 + d^2)) + (6*b*c^5*tan(1/2*f*x + 1/2*e)^5 - 18*a*c^4*d*tan(1/2*f*x + 1/2*e)^5 - 6*b*c^4*d*tan(1/2*f*x + 1/
2*e)^5 + 27*a*c^3*d^2*tan(1/2*f*x + 1/2*e)^5 + 12*b*c^3*d^2*tan(1/2*f*x + 1/2*e)^5 - 6*a*c^2*d^3*tan(1/2*f*x +
1/2*e)^5 - 27*b*c^2*d^3*tan(1/2*f*x + 1/2*e)^5 + 3*a*c*d^4*tan(1/2*f*x + 1/2*e)^5 + 12*b*c*d^4*tan(1/2*f*x +
1/2*e)^5 - 6*a*d^5*tan(1/2*f*x + 1/2*e)^5 + 3*b*d^5*tan(1/2*f*x + 1/2*e)^5 - 12*b*c^5*tan(1/2*f*x + 1/2*e)^3 +
36*a*c^4*d*tan(1/2*f*x + 1/2*e)^3 - 16*b*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 - 32*a*c^2*d^3*tan(1/2*f*x + 1/2*e)^3
+ 28*b*c*d^4*tan(1/2*f*x + 1/2*e)^3 - 4*a*d^5*tan(1/2*f*x + 1/2*e)^3 + 6*b*c^5*tan(1/2*f*x + 1/2*e) - 18*a*c^
4*d*tan(1/2*f*x + 1/2*e) + 6*b*c^4*d*tan(1/2*f*x + 1/2*e) - 27*a*c^3*d^2*tan(1/2*f*x + 1/2*e) + 12*b*c^3*d^2*t
an(1/2*f*x + 1/2*e) - 6*a*c^2*d^3*tan(1/2*f*x + 1/2*e) + 27*b*c^2*d^3*tan(1/2*f*x + 1/2*e) - 3*a*c*d^4*tan(1/2
*f*x + 1/2*e) + 12*b*c*d^4*tan(1/2*f*x + 1/2*e) - 6*a*d^5*tan(1/2*f*x + 1/2*e) - 3*b*d^5*tan(1/2*f*x + 1/2*e))
/((c^6 - 3*c^4*d^2 + 3*c^2*d^4 - d^6)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)^3))/f