3.226 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^5}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=287 \[ -\frac{2 d \left (72 c^2 d^2+15 c^3 d+2 c^4-180 c d^3+76 d^4\right ) \tan (e+f x)}{15 a^3 f}+\frac{d^3 \left (20 c^2-30 c d+13 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{2 a^3 f}+\frac{(c-d) \left (2 c^2+15 c d+76 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{15 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{d^2 \left (30 c^2 d+4 c^3+146 c d^2-195 d^3\right ) \tan (e+f x) \sec (e+f x)}{30 a^3 f}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^4}{5 f (a \sec (e+f x)+a)^3}+\frac{(c-d) (2 c+11 d) \tan (e+f x) (c+d \sec (e+f x))^3}{15 a f (a \sec (e+f x)+a)^2} \]

[Out]

(d^3*(20*c^2 - 30*c*d + 13*d^2)*ArcTanh[Sin[e + f*x]])/(2*a^3*f) - (2*d*(2*c^4 + 15*c^3*d + 72*c^2*d^2 - 180*c
*d^3 + 76*d^4)*Tan[e + f*x])/(15*a^3*f) - (d^2*(4*c^3 + 30*c^2*d + 146*c*d^2 - 195*d^3)*Sec[e + f*x]*Tan[e + f
*x])/(30*a^3*f) + ((c - d)*(2*c^2 + 15*c*d + 76*d^2)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec
[e + f*x])) + ((c - d)*(2*c + 11*d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(15*a*f*(a + a*Sec[e + f*x])^2) + ((c
- d)*(c + d*Sec[e + f*x])^4*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3)

________________________________________________________________________________________

Rubi [A]  time = 0.416137, antiderivative size = 329, normalized size of antiderivative = 1.15, 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 = {3987, 98, 150, 147, 63, 217, 203} \[ \frac{d^3 \left (20 c^2-30 c d+13 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(c-d) \left (2 c^2+15 c d+76 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{15 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac{d \tan (e+f x) \left (d \left (30 c^2 d+4 c^3+146 c d^2-195 d^3\right ) \sec (e+f x)+4 \left (72 c^2 d^2+15 c^3 d+2 c^4-180 c d^3+76 d^4\right )\right )}{30 a^3 f}+\frac{(c-d) \tan (e+f x) (c+d \sec (e+f x))^4}{5 f (a \sec (e+f x)+a)^3}+\frac{(c-d) (2 c+11 d) \tan (e+f x) (c+d \sec (e+f x))^3}{15 a f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^5)/(a + a*Sec[e + f*x])^3,x]

[Out]

(d^3*(20*c^2 - 30*c*d + 13*d^2)*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(a^2
*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + ((c - d)*(2*c^2 + 15*c*d + 76*d^2)*(c + d*Sec[e + f*x]
)^2*Tan[e + f*x])/(15*f*(a^3 + a^3*Sec[e + f*x])) + ((c - d)*(2*c + 11*d)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])
/(15*a*f*(a + a*Sec[e + f*x])^2) + ((c - d)*(c + d*Sec[e + f*x])^4*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3)
- (d*(4*(2*c^4 + 15*c^3*d + 72*c^2*d^2 - 180*c*d^3 + 76*d^4) + d*(4*c^3 + 30*c^2*d + 146*c*d^2 - 195*d^3)*Sec[
e + f*x])*Tan[e + f*x])/(30*a^3*f)

Rule 3987

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[(a^2*g*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x
]]), Subst[Int[((g*x)^(p - 1)*(a + b*x)^(m - 1/2)*(c + d*x)^n)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; Free
Q[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p,
1] || IntegerQ[m - 1/2])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 150

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (e+f x) (c+d \sec (e+f x))^5}{(a+a \sec (e+f x))^3} \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^5}{\sqrt{a-a x} (a+a x)^{7/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^3 \left (-a^2 (2 c-d) (c+4 d)+a^2 (2 c-7 d) d x\right )}{\sqrt{a-a x} (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) (2 c+11 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x)^2 \left (-a^4 \left (2 c^3+9 c^2 d+37 c d^2-33 d^3\right )+a^4 d \left (4 c^2+24 c d-43 d^2\right ) x\right )}{\sqrt{a-a x} (a+a x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) \left (2 c^2+15 c d+76 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+11 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c+d x) \left (-a^6 d^2 \left (2 c^2+165 c d-152 d^2\right )+a^6 d \left (4 c^3+30 c^2 d+146 c d^2-195 d^3\right ) x\right )}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{15 a^7 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) \left (2 c^2+15 c d+76 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+11 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^4+15 c^3 d+72 c^2 d^2-180 c d^3+76 d^4\right )+d \left (4 c^3+30 c^2 d+146 c d^2-195 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}-\frac{\left (d^3 \left (20 c^2-30 c d+13 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-a x} \sqrt{a+a x}} \, dx,x,\sec (e+f x)\right )}{2 a f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) \left (2 c^2+15 c d+76 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+11 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^4+15 c^3 d+72 c^2 d^2-180 c d^3+76 d^4\right )+d \left (4 c^3+30 c^2 d+146 c d^2-195 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}+\frac{\left (d^3 \left (20 c^2-30 c d+13 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2 a-x^2}} \, dx,x,\sqrt{a-a \sec (e+f x)}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(c-d) \left (2 c^2+15 c d+76 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+11 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^4+15 c^3 d+72 c^2 d^2-180 c d^3+76 d^4\right )+d \left (4 c^3+30 c^2 d+146 c d^2-195 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}+\frac{\left (d^3 \left (20 c^2-30 c d+13 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right )}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{d^3 \left (20 c^2-30 c d+13 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a+a \sec (e+f x)}}\right ) \tan (e+f x)}{a^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{(c-d) \left (2 c^2+15 c d+76 d^2\right ) (c+d \sec (e+f x))^2 \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )}+\frac{(c-d) (2 c+11 d) (c+d \sec (e+f x))^3 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac{(c-d) (c+d \sec (e+f x))^4 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac{d \left (4 \left (2 c^4+15 c^3 d+72 c^2 d^2-180 c d^3+76 d^4\right )+d \left (4 c^3+30 c^2 d+146 c d^2-195 d^3\right ) \sec (e+f x)\right ) \tan (e+f x)}{30 a^3 f}\\ \end{align*}

Mathematica [A]  time = 2.14813, size = 439, normalized size = 1.53 \[ \frac{2 \sin \left (\frac{1}{2} (e+f x)\right ) \cos \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (120 c^3 d^2 \cos (3 (e+f x))+20 c^3 d^2 \cos (4 (e+f x))-1020 c^2 d^3 \cos (3 (e+f x))-220 c^2 d^3 \cos (4 (e+f x))+3 \left (120 c^3 d^2-1020 c^2 d^3+90 c^4 d+12 c^5+1910 c d^4-777 d^5\right ) \cos (e+f x)+6 \left (60 c^3 d^2-360 c^2 d^3+20 c^4 d+6 c^5+630 c d^4-261 d^5\right ) \cos (2 (e+f x))+340 c^3 d^2-1940 c^2 d^3+90 c^4 d \cos (3 (e+f x))+15 c^4 d \cos (4 (e+f x))+105 c^4 d+12 c^5 \cos (3 (e+f x))+7 c^5 \cos (4 (e+f x))+29 c^5+1710 c d^4 \cos (3 (e+f x))+360 c d^4 \cos (4 (e+f x))+3420 c d^4-717 d^5 \cos (3 (e+f x))-152 d^5 \cos (4 (e+f x))-1354 d^5\right )-480 d^3 \left (20 c^2-30 c d+13 d^2\right ) \cos ^6\left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{120 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^5)/(a + a*Sec[e + f*x])^3,x]

[Out]

(-480*d^3*(20*c^2 - 30*c*d + 13*d^2)*Cos[(e + f*x)/2]^6*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2]]) + 2*Cos[(e + f*x)/2]*(29*c^5 + 105*c^4*d + 340*c^3*d^2 - 1940*c^2*d^3 + 3420*c
*d^4 - 1354*d^5 + 3*(12*c^5 + 90*c^4*d + 120*c^3*d^2 - 1020*c^2*d^3 + 1910*c*d^4 - 777*d^5)*Cos[e + f*x] + 6*(
6*c^5 + 20*c^4*d + 60*c^3*d^2 - 360*c^2*d^3 + 630*c*d^4 - 261*d^5)*Cos[2*(e + f*x)] + 12*c^5*Cos[3*(e + f*x)]
+ 90*c^4*d*Cos[3*(e + f*x)] + 120*c^3*d^2*Cos[3*(e + f*x)] - 1020*c^2*d^3*Cos[3*(e + f*x)] + 1710*c*d^4*Cos[3*
(e + f*x)] - 717*d^5*Cos[3*(e + f*x)] + 7*c^5*Cos[4*(e + f*x)] + 15*c^4*d*Cos[4*(e + f*x)] + 20*c^3*d^2*Cos[4*
(e + f*x)] - 220*c^2*d^3*Cos[4*(e + f*x)] + 360*c*d^4*Cos[4*(e + f*x)] - 152*d^5*Cos[4*(e + f*x)])*Sec[e + f*x
]^2*Sin[(e + f*x)/2])/(120*a^3*f*(1 + Cos[e + f*x])^3)

________________________________________________________________________________________

Maple [B]  time = 0.104, size = 679, normalized size = 2.4 \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))^5/(a+a*sec(f*x+e))^3,x)

[Out]

1/20/f*c^5/a^3*tan(1/2*f*x+1/2*e)^5-1/6/f*c^5/a^3*tan(1/2*f*x+1/2*e)^3+1/4/f*c^5/a^3*tan(1/2*f*x+1/2*e)+7/2/f/
a^3*d^5/(tan(1/2*f*x+1/2*e)+1)+1/2/f/a^3*d^5/(tan(1/2*f*x+1/2*e)-1)^2-13/2/f/a^3*ln(tan(1/2*f*x+1/2*e)-1)*d^5+
7/2/f/a^3*d^5/(tan(1/2*f*x+1/2*e)-1)-1/20/f/a^3*tan(1/2*f*x+1/2*e)^5*d^5-2/3/f/a^3*tan(1/2*f*x+1/2*e)^3*d^5-31
/4/f/a^3*tan(1/2*f*x+1/2*e)*d^5-1/2/f/a^3*d^5/(tan(1/2*f*x+1/2*e)+1)^2+13/2/f/a^3*ln(tan(1/2*f*x+1/2*e)+1)*d^5
-10/f/a^3*ln(tan(1/2*f*x+1/2*e)-1)*c^2*d^3-1/2/f/a^3*tan(1/2*f*x+1/2*e)^5*c^2*d^3+1/4/f/a^3*tan(1/2*f*x+1/2*e)
^5*c*d^4-35/2/f/a^3*tan(1/2*f*x+1/2*e)*c^2*d^3+85/4/f/a^3*tan(1/2*f*x+1/2*e)*c*d^4+10/f/a^3*ln(tan(1/2*f*x+1/2
*e)+1)*c^2*d^3-15/f/a^3*ln(tan(1/2*f*x+1/2*e)+1)*c*d^4-5/f/a^3*d^4/(tan(1/2*f*x+1/2*e)+1)*c-1/4/f/a^3*tan(1/2*
f*x+1/2*e)^5*c^4*d+1/2/f/a^3*tan(1/2*f*x+1/2*e)^5*c^3*d^2+15/f/a^3*ln(tan(1/2*f*x+1/2*e)-1)*c*d^4-5/f/a^3*d^4/
(tan(1/2*f*x+1/2*e)-1)*c+5/3/f/a^3*tan(1/2*f*x+1/2*e)^3*c^3*d^2-10/3/f/a^3*tan(1/2*f*x+1/2*e)^3*c^2*d^3+5/2/f/
a^3*tan(1/2*f*x+1/2*e)^3*c*d^4+5/4/f/a^3*tan(1/2*f*x+1/2*e)*c^4*d+5/2/f/a^3*tan(1/2*f*x+1/2*e)*c^3*d^2

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Maxima [B]  time = 1.08155, size = 930, normalized size = 3.24 \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))^5/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/60*(d^5*(60*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 7*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/(a^3 - 2*a^3*sin(f*
x + e)^2/(cos(f*x + e) + 1)^2 + a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4) + (465*sin(f*x + e)/(cos(f*x + e) + 1
) + 40*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 390*log(sin(f*x + e)
/(cos(f*x + e) + 1) + 1)/a^3 + 390*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) - 15*c*d^4*(40*sin(f*x + e)/(
(a^3 - a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)) + (85*sin(f*x + e)/(cos(f*x + e) + 1) + 10
*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 60*log(sin(f*x + e)/(cos(f*x
+ e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) + 10*c^2*d^3*((105*sin(f*x + e)/(cos(f*
x + e) + 1) + 20*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 - 60*log(sin
(f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) - 10*c^3*d^2*(15*sin(
f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/
a^3 - c^5*(15*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos
(f*x + e) + 1)^5)/a^3 - 15*c^4*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3
)/f

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Fricas [A]  time = 0.57135, size = 1200, normalized size = 4.18 \begin{align*} \frac{15 \,{\left ({\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{5} + 3 \,{\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{4} + 3 \,{\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{3} +{\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left ({\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{5} + 3 \,{\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{4} + 3 \,{\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{3} +{\left (20 \, c^{2} d^{3} - 30 \, c d^{4} + 13 \, d^{5}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (15 \, d^{5} + 2 \,{\left (7 \, c^{5} + 15 \, c^{4} d + 20 \, c^{3} d^{2} - 220 \, c^{2} d^{3} + 360 \, c d^{4} - 152 \, d^{5}\right )} \cos \left (f x + e\right )^{4} + 3 \,{\left (4 \, c^{5} + 30 \, c^{4} d + 40 \, c^{3} d^{2} - 340 \, c^{2} d^{3} + 570 \, c d^{4} - 239 \, d^{5}\right )} \cos \left (f x + e\right )^{3} +{\left (4 \, c^{5} + 30 \, c^{4} d + 140 \, c^{3} d^{2} - 640 \, c^{2} d^{3} + 1170 \, c d^{4} - 479 \, d^{5}\right )} \cos \left (f x + e\right )^{2} + 15 \,{\left (10 \, c d^{4} - 3 \, d^{5}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{60 \,{\left (a^{3} f \cos \left (f x + e\right )^{5} + 3 \, a^{3} f \cos \left (f x + e\right )^{4} + 3 \, a^{3} f \cos \left (f x + e\right )^{3} + a^{3} f \cos \left (f x + e\right )^{2}\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))^5/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/60*(15*((20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(f*x + e)^5 + 3*(20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(f*x + e)^4
+ 3*(20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(f*x + e)^3 + (20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(f*x + e)^2)*log(sin
(f*x + e) + 1) - 15*((20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(f*x + e)^5 + 3*(20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(
f*x + e)^4 + 3*(20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(f*x + e)^3 + (20*c^2*d^3 - 30*c*d^4 + 13*d^5)*cos(f*x + e)
^2)*log(-sin(f*x + e) + 1) + 2*(15*d^5 + 2*(7*c^5 + 15*c^4*d + 20*c^3*d^2 - 220*c^2*d^3 + 360*c*d^4 - 152*d^5)
*cos(f*x + e)^4 + 3*(4*c^5 + 30*c^4*d + 40*c^3*d^2 - 340*c^2*d^3 + 570*c*d^4 - 239*d^5)*cos(f*x + e)^3 + (4*c^
5 + 30*c^4*d + 140*c^3*d^2 - 640*c^2*d^3 + 1170*c*d^4 - 479*d^5)*cos(f*x + e)^2 + 15*(10*c*d^4 - 3*d^5)*cos(f*
x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^5 + 3*a^3*f*cos(f*x + e)^4 + 3*a^3*f*cos(f*x + e)^3 + a^3*f*cos(f*x
+ e)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c^{5} \sec{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{d^{5} \sec ^{6}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{5 c d^{4} \sec ^{5}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{10 c^{2} d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{10 c^{3} d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx + \int \frac{5 c^{4} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec{\left (e + f x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**5/(a+a*sec(f*x+e))**3,x)

[Out]

(Integral(c**5*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(d**5*sec
(e + f*x)**6/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(5*c*d**4*sec(e + f*x)**
5/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(10*c**2*d**3*sec(e + f*x)**4/(sec(
e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(10*c**3*d**2*sec(e + f*x)**3/(sec(e + f*x
)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(5*c**4*d*sec(e + f*x)**2/(sec(e + f*x)**3 + 3*se
c(e + f*x)**2 + 3*sec(e + f*x) + 1), x))/a**3

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Giac [A]  time = 1.30357, size = 713, normalized size = 2.48 \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))^5/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/60*(30*(20*c^2*d^3 - 30*c*d^4 + 13*d^5)*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 30*(20*c^2*d^3 - 30*c*d^4 +
13*d^5)*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 - 60*(10*c*d^4*tan(1/2*f*x + 1/2*e)^3 - 7*d^5*tan(1/2*f*x + 1/
2*e)^3 - 10*c*d^4*tan(1/2*f*x + 1/2*e) + 5*d^5*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1)^2*a^3) + (3
*a^12*c^5*tan(1/2*f*x + 1/2*e)^5 - 15*a^12*c^4*d*tan(1/2*f*x + 1/2*e)^5 + 30*a^12*c^3*d^2*tan(1/2*f*x + 1/2*e)
^5 - 30*a^12*c^2*d^3*tan(1/2*f*x + 1/2*e)^5 + 15*a^12*c*d^4*tan(1/2*f*x + 1/2*e)^5 - 3*a^12*d^5*tan(1/2*f*x +
1/2*e)^5 - 10*a^12*c^5*tan(1/2*f*x + 1/2*e)^3 + 100*a^12*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 - 200*a^12*c^2*d^3*tan
(1/2*f*x + 1/2*e)^3 + 150*a^12*c*d^4*tan(1/2*f*x + 1/2*e)^3 - 40*a^12*d^5*tan(1/2*f*x + 1/2*e)^3 + 15*a^12*c^5
*tan(1/2*f*x + 1/2*e) + 75*a^12*c^4*d*tan(1/2*f*x + 1/2*e) + 150*a^12*c^3*d^2*tan(1/2*f*x + 1/2*e) - 1050*a^12
*c^2*d^3*tan(1/2*f*x + 1/2*e) + 1275*a^12*c*d^4*tan(1/2*f*x + 1/2*e) - 465*a^12*d^5*tan(1/2*f*x + 1/2*e))/a^15
)/f