3.194 \(\int \sec (e+f x) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=242 \[ -\frac{a^2 \left (-44 c^2 d^2-10 c^3 d+c^4-40 c d^3-12 d^4\right ) \tan (e+f x)}{10 d f}+\frac{3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{a^2 \left (c^2-10 c d-12 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{20 d f}-\frac{a^2 \left (-20 c^2 d+2 c^3-57 c d^2-30 d^3\right ) \tan (e+f x) \sec (e+f x)}{40 f}+\frac{a^2 \tan (e+f x) (c+d \sec (e+f x))^4}{5 d f}-\frac{a^2 (c-10 d) \tan (e+f x) (c+d \sec (e+f x))^3}{20 d f} \]

[Out]

(3*a^2*(2*c + d)*(2*c^2 + 3*c*d + 2*d^2)*ArcTanh[Sin[e + f*x]])/(8*f) - (a^2*(c^4 - 10*c^3*d - 44*c^2*d^2 - 40
*c*d^3 - 12*d^4)*Tan[e + f*x])/(10*d*f) - (a^2*(2*c^3 - 20*c^2*d - 57*c*d^2 - 30*d^3)*Sec[e + f*x]*Tan[e + f*x
])/(40*f) - (a^2*(c^2 - 10*c*d - 12*d^2)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(20*d*f) - (a^2*(c - 10*d)*(c +
d*Sec[e + f*x])^3*Tan[e + f*x])/(20*d*f) + (a^2*(c + d*Sec[e + f*x])^4*Tan[e + f*x])/(5*d*f)

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Rubi [A]  time = 0.342701, antiderivative size = 277, normalized size of antiderivative = 1.14, 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, 100, 147, 50, 63, 217, 203} \[ \frac{3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac{3 a^3 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{8 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right )}{20 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))^2}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*(2*c + d)*(2*c^2 + 3*c*d + 2*d^2)*Tan[e + f*x])/(8*f) + (3*a^3*(2*c + d)*(2*c^2 + 3*c*d + 2*d^2)*ArcTan
[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(4*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*S
ec[e + f*x]]) + ((2*c + d)*(2*c^2 + 3*c*d + 2*d^2)*(a^2 + a^2*Sec[e + f*x])*Tan[e + f*x])/(8*f) + (d*(a + a*Se
c[e + f*x])^2*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(5*f) + (d*(a + a*Sec[e + f*x])^2*(2*(8*c^2 + 5*c*d + 2*d^2
) + d*(7*c + 2*d)*Sec[e + f*x])*Tan[e + f*x])/(20*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 100

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 \sec (e+f x) (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^3 \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{3/2} (c+d x)^3}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(a+a x)^{3/2} (c+d x) \left (-a^2 \left (5 c^2+2 c d+2 d^2\right )-a^2 d (7 c+2 d) x\right )}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{5 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac{d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}-\frac{\left (a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{3/2}}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac{d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}-\frac{\left (3 a^3 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+a x}}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{8 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac{(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac{d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}-\frac{\left (3 a^4 (2 c+d) \left (2 c^2+3 c d+2 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 )}{8 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac{(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac{d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}+\frac{\left (3 a^3 (2 c+d) \left (2 c^2+3 c d+2 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 )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac{(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac{d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}+\frac{\left (3 a^3 (2 c+d) \left (2 c^2+3 c d+2 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 )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{3 a^2 (2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \tan (e+f x)}{8 f}+\frac{3 a^3 (2 c+d) \left (2 c^2+3 c d+2 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)}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{(2 c+d) \left (2 c^2+3 c d+2 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{8 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f}+\frac{d (a+a \sec (e+f x))^2 \left (2 \left (8 c^2+5 c d+2 d^2\right )+d (7 c+2 d) \sec (e+f x)\right ) \tan (e+f x)}{20 f}\\ \end{align*}

Mathematica [A]  time = 1.45491, size = 326, normalized size = 1.35 \[ -\frac{a^2 (\cos (e+f x)+1)^2 \sec ^4\left (\frac{1}{2} (e+f x)\right ) \sec ^5(e+f x) \left (120 \left (8 c^2 d+4 c^3+7 c d^2+2 d^3\right ) \cos ^5(e+f x) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-2 \sin (e+f x) \left (5 \left (72 c^2 d+12 c^3+87 c d^2+34 d^3\right ) \cos (e+f x)+16 \left (30 c^2 d+10 c^3+30 c d^2+9 d^3\right ) \cos (2 (e+f x))+120 c^2 d \cos (3 (e+f x))+100 c^2 d \cos (4 (e+f x))+380 c^2 d+20 c^3 \cos (3 (e+f x))+40 c^3 \cos (4 (e+f x))+120 c^3+105 c d^2 \cos (3 (e+f x))+80 c d^2 \cos (4 (e+f x))+400 c d^2+30 d^3 \cos (3 (e+f x))+24 d^3 \cos (4 (e+f x))+152 d^3\right )\right )}{1280 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(1 + Cos[e + f*x])^2*Sec[(e + f*x)/2]^4*Sec[e + f*x]^5*(120*(4*c^3 + 8*c^2*d + 7*c*d^2 + 2*d^3)*Cos[e +
f*x]^5*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - 2*(120*c^3 + 38
0*c^2*d + 400*c*d^2 + 152*d^3 + 5*(12*c^3 + 72*c^2*d + 87*c*d^2 + 34*d^3)*Cos[e + f*x] + 16*(10*c^3 + 30*c^2*d
+ 30*c*d^2 + 9*d^3)*Cos[2*(e + f*x)] + 20*c^3*Cos[3*(e + f*x)] + 120*c^2*d*Cos[3*(e + f*x)] + 105*c*d^2*Cos[3
*(e + f*x)] + 30*d^3*Cos[3*(e + f*x)] + 40*c^3*Cos[4*(e + f*x)] + 100*c^2*d*Cos[4*(e + f*x)] + 80*c*d^2*Cos[4*
(e + f*x)] + 24*d^3*Cos[4*(e + f*x)])*Sin[e + f*x]))/(1280*f)

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Maple [A]  time = 0.063, size = 420, normalized size = 1.7 \begin{align*}{\frac{3\,{a}^{2}{c}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+5\,{\frac{{a}^{2}{c}^{2}d\tan \left ( fx+e \right ) }{f}}+{\frac{21\,{a}^{2}{d}^{2}c\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{21\,{a}^{2}{d}^{2}c\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{6\,{a}^{2}{d}^{3}\tan \left ( fx+e \right ) }{5\,f}}+{\frac{3\,{a}^{2}{d}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{5\,f}}+2\,{\frac{{a}^{2}{c}^{3}\tan \left ( fx+e \right ) }{f}}+3\,{\frac{{a}^{2}{c}^{2}d\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{f}}+3\,{\frac{{a}^{2}{c}^{2}d\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+4\,{\frac{{a}^{2}{d}^{2}c\tan \left ( fx+e \right ) }{f}}+2\,{\frac{{a}^{2}{d}^{2}c\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{{a}^{2}{d}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{2\,f}}+{\frac{3\,{a}^{2}{d}^{3}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{4\,f}}+{\frac{3\,{a}^{2}{d}^{3}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{4\,f}}+{\frac{{a}^{2}{c}^{3}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{{a}^{2}{c}^{2}d\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{3\,{a}^{2}{d}^{2}c\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{{a}^{2}{d}^{3}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5\,f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^3,x)

[Out]

3/2/f*a^2*c^3*ln(sec(f*x+e)+tan(f*x+e))+5/f*a^2*c^2*d*tan(f*x+e)+21/8/f*a^2*d^2*c*sec(f*x+e)*tan(f*x+e)+21/8/f
*a^2*d^2*c*ln(sec(f*x+e)+tan(f*x+e))+6/5/f*a^2*d^3*tan(f*x+e)+3/5/f*a^2*d^3*tan(f*x+e)*sec(f*x+e)^2+2/f*a^2*c^
3*tan(f*x+e)+3/f*a^2*c^2*d*sec(f*x+e)*tan(f*x+e)+3/f*a^2*c^2*d*ln(sec(f*x+e)+tan(f*x+e))+4/f*a^2*d^2*c*tan(f*x
+e)+2/f*a^2*d^2*c*tan(f*x+e)*sec(f*x+e)^2+1/2/f*a^2*d^3*tan(f*x+e)*sec(f*x+e)^3+3/4/f*a^2*d^3*sec(f*x+e)*tan(f
*x+e)+3/4/f*a^2*d^3*ln(sec(f*x+e)+tan(f*x+e))+1/2*a^2*c^3*sec(f*x+e)*tan(f*x+e)/f+1/f*a^2*c^2*d*tan(f*x+e)*sec
(f*x+e)^2+3/4/f*a^2*d^2*c*tan(f*x+e)*sec(f*x+e)^3+1/5/f*a^2*d^3*tan(f*x+e)*sec(f*x+e)^4

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Maxima [B]  time = 0.988417, size = 633, normalized size = 2.62 \begin{align*} \frac{240 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c^{2} d + 480 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c d^{2} + 16 \,{\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{2} d^{3} + 80 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} d^{3} - 45 \, a^{2} c d^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 30 \, a^{2} d^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 60 \, a^{2} c^{3}{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 360 \, a^{2} c^{2} d{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 180 \, a^{2} c d^{2}{\left (\frac{2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 240 \, a^{2} c^{3} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 480 \, a^{2} c^{3} \tan \left (f x + e\right ) + 720 \, a^{2} c^{2} d \tan \left (f x + e\right )}{240 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

1/240*(240*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c^2*d + 480*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c*d^2 + 16*
(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^2*d^3 + 80*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*d^
3 - 45*a^2*c*d^2*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(sin(f*
x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 30*a^2*d^3*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2
*sin(f*x + e)^2 + 1) - 3*log(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 60*a^2*c^3*(2*sin(f*x + e)/(sin(f*
x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 360*a^2*c^2*d*(2*sin(f*x + e)/(sin(f*x + e)^2
- 1) - log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) - 180*a^2*c*d^2*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) -
log(sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) + 240*a^2*c^3*log(sec(f*x + e) + tan(f*x + e)) + 480*a^2*c^3*ta
n(f*x + e) + 720*a^2*c^2*d*tan(f*x + e))/f

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Fricas [A]  time = 0.530885, size = 667, normalized size = 2.76 \begin{align*} \frac{15 \,{\left (4 \, a^{2} c^{3} + 8 \, a^{2} c^{2} d + 7 \, a^{2} c d^{2} + 2 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left (4 \, a^{2} c^{3} + 8 \, a^{2} c^{2} d + 7 \, a^{2} c d^{2} + 2 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (8 \, a^{2} d^{3} + 8 \,{\left (10 \, a^{2} c^{3} + 25 \, a^{2} c^{2} d + 20 \, a^{2} c d^{2} + 6 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{4} + 5 \,{\left (4 \, a^{2} c^{3} + 24 \, a^{2} c^{2} d + 21 \, a^{2} c d^{2} + 6 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{3} + 8 \,{\left (5 \, a^{2} c^{2} d + 10 \, a^{2} c d^{2} + 3 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )^{2} + 10 \,{\left (3 \, a^{2} c d^{2} + 2 \, a^{2} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{80 \, f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

1/80*(15*(4*a^2*c^3 + 8*a^2*c^2*d + 7*a^2*c*d^2 + 2*a^2*d^3)*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 15*(4*a^2*
c^3 + 8*a^2*c^2*d + 7*a^2*c*d^2 + 2*a^2*d^3)*cos(f*x + e)^5*log(-sin(f*x + e) + 1) + 2*(8*a^2*d^3 + 8*(10*a^2*
c^3 + 25*a^2*c^2*d + 20*a^2*c*d^2 + 6*a^2*d^3)*cos(f*x + e)^4 + 5*(4*a^2*c^3 + 24*a^2*c^2*d + 21*a^2*c*d^2 + 6
*a^2*d^3)*cos(f*x + e)^3 + 8*(5*a^2*c^2*d + 10*a^2*c*d^2 + 3*a^2*d^3)*cos(f*x + e)^2 + 10*(3*a^2*c*d^2 + 2*a^2
*d^3)*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**2*(c+d*sec(f*x+e))**3,x)

[Out]

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

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Giac [B]  time = 1.47218, size = 714, normalized size = 2.95 \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+a*sec(f*x+e))^2*(c+d*sec(f*x+e))^3,x, algorithm="giac")

[Out]

1/40*(15*(4*a^2*c^3 + 8*a^2*c^2*d + 7*a^2*c*d^2 + 2*a^2*d^3)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15*(4*a^2*c^
3 + 8*a^2*c^2*d + 7*a^2*c*d^2 + 2*a^2*d^3)*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(60*a^2*c^3*tan(1/2*f*x + 1/
2*e)^9 + 120*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^9 + 105*a^2*c*d^2*tan(1/2*f*x + 1/2*e)^9 + 30*a^2*d^3*tan(1/2*f*x
+ 1/2*e)^9 - 280*a^2*c^3*tan(1/2*f*x + 1/2*e)^7 - 560*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^7 - 490*a^2*c*d^2*tan(1/2
*f*x + 1/2*e)^7 - 140*a^2*d^3*tan(1/2*f*x + 1/2*e)^7 + 480*a^2*c^3*tan(1/2*f*x + 1/2*e)^5 + 1120*a^2*c^2*d*tan
(1/2*f*x + 1/2*e)^5 + 800*a^2*c*d^2*tan(1/2*f*x + 1/2*e)^5 + 288*a^2*d^3*tan(1/2*f*x + 1/2*e)^5 - 360*a^2*c^3*
tan(1/2*f*x + 1/2*e)^3 - 1040*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 790*a^2*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 180*a^
2*d^3*tan(1/2*f*x + 1/2*e)^3 + 100*a^2*c^3*tan(1/2*f*x + 1/2*e) + 360*a^2*c^2*d*tan(1/2*f*x + 1/2*e) + 375*a^2
*c*d^2*tan(1/2*f*x + 1/2*e) + 130*a^2*d^3*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^5)/f