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

Optimal. Leaf size=257 \[ \frac{a^3 \left (72 c^2 d^2-15 c^3 d+2 c^4+180 c d^3+76 d^4\right ) \tan (e+f x)}{30 d^2 f}+\frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}+\frac{a^3 \left (2 c^2-15 c d+76 d^2\right ) \tan (e+f x) (c+d \sec (e+f x))^2}{60 d^2 f}+\frac{a^3 \left (-30 c^2 d+4 c^3+146 c d^2+195 d^3\right ) \tan (e+f x) \sec (e+f x)}{120 d f}-\frac{a^3 (2 c-11 d) \tan (e+f x) (c+d \sec (e+f x))^3}{20 d^2 f}+\frac{\tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right ) (c+d \sec (e+f x))^3}{5 d f} \]

[Out]

(a^3*(20*c^2 + 30*c*d + 13*d^2)*ArcTanh[Sin[e + f*x]])/(8*f) + (a^3*(2*c^4 - 15*c^3*d + 72*c^2*d^2 + 180*c*d^3
+ 76*d^4)*Tan[e + f*x])/(30*d^2*f) + (a^3*(4*c^3 - 30*c^2*d + 146*c*d^2 + 195*d^3)*Sec[e + f*x]*Tan[e + f*x])
/(120*d*f) + (a^3*(2*c^2 - 15*c*d + 76*d^2)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(60*d^2*f) - (a^3*(2*c - 11*d
)*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(20*d^2*f) + ((a^3 + a^3*Sec[e + f*x])*(c + d*Sec[e + f*x])^3*Tan[e + f
*x])/(5*d*f)

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Rubi [A]  time = 0.299609, antiderivative size = 273, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3987, 90, 80, 50, 63, 217, 203} \[ \frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x)}{8 f}+\frac{a^4 \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{\left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{24 f}+\frac{a \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x) (a \sec (e+f x)+a)^2}{60 f}+\frac{3 d (2 c+d) \tan (e+f x) (a \sec (e+f x)+a)^3}{20 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^3 (c+d \sec (e+f x))}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(20*c^2 + 30*c*d + 13*d^2)*Tan[e + f*x])/(8*f) + (a^4*(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])/(4*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (a
*(20*c^2 + 30*c*d + 13*d^2)*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(60*f) + (3*d*(2*c + d)*(a + a*Sec[e + f*x])^
3*Tan[e + f*x])/(20*f) + ((20*c^2 + 30*c*d + 13*d^2)*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(24*f) + (d*(a + a
*Sec[e + f*x])^3*(c + d*Sec[e + f*x])*Tan[e + f*x])/(5*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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
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))^3 (c+d \sec (e+f x))^2 \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{5/2} (c+d x)^2}{\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))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(a+a x)^{5/2} \left (-a^2 \left (5 c^2+3 c d+d^2\right )-3 a^2 d (2 c+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{3 d (2 c+d) (a+a \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{d (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}-\frac{\left (a^2 \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{5/2}}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{20 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{a \left (20 c^2+30 c d+13 d^2\right ) (a+a \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{3 d (2 c+d) (a+a \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{d (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}-\frac{\left (a^3 \left (20 c^2+30 c d+13 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 )}{12 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{a \left (20 c^2+30 c d+13 d^2\right ) (a+a \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{3 d (2 c+d) (a+a \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{\left (20 c^2+30 c d+13 d^2\right ) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}-\frac{\left (a^4 \left (20 c^2+30 c d+13 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{a^3 \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x)}{8 f}+\frac{a \left (20 c^2+30 c d+13 d^2\right ) (a+a \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{3 d (2 c+d) (a+a \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{\left (20 c^2+30 c d+13 d^2\right ) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}-\frac{\left (a^5 \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 )}{8 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x)}{8 f}+\frac{a \left (20 c^2+30 c d+13 d^2\right ) (a+a \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{3 d (2 c+d) (a+a \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{\left (20 c^2+30 c d+13 d^2\right ) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}+\frac{\left (a^4 \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 )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x)}{8 f}+\frac{a \left (20 c^2+30 c d+13 d^2\right ) (a+a \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{3 d (2 c+d) (a+a \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{\left (20 c^2+30 c d+13 d^2\right ) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}+\frac{\left (a^4 \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 )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \tan (e+f x)}{8 f}+\frac{a^4 \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)}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}+\frac{a \left (20 c^2+30 c d+13 d^2\right ) (a+a \sec (e+f x))^2 \tan (e+f x)}{60 f}+\frac{3 d (2 c+d) (a+a \sec (e+f x))^3 \tan (e+f x)}{20 f}+\frac{\left (20 c^2+30 c d+13 d^2\right ) \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^3 (c+d \sec (e+f x)) \tan (e+f x)}{5 f}\\ \end{align*}

Mathematica [A]  time = 2.5544, size = 433, normalized size = 1.68 \[ -\frac{a^3 (\cos (e+f x)+1)^3 \sec ^6\left (\frac{1}{2} (e+f x)\right ) \sec ^5(e+f x) \left (240 \left (20 c^2+30 c d+13 d^2\right ) \cos ^5(e+f x) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-\sec (e) \left (-240 \left (7 c^2+10 c d+3 d^2\right ) \sin (2 e+f x)+80 \left (34 c^2+60 c d+29 d^2\right ) \sin (f x)+360 c^2 \sin (e+2 f x)+360 c^2 \sin (3 e+2 f x)+1840 c^2 \sin (2 e+3 f x)-360 c^2 \sin (4 e+3 f x)+180 c^2 \sin (3 e+4 f x)+180 c^2 \sin (5 e+4 f x)+440 c^2 \sin (4 e+5 f x)+1140 c d \sin (e+2 f x)+1140 c d \sin (3 e+2 f x)+3360 c d \sin (2 e+3 f x)-240 c d \sin (4 e+3 f x)+450 c d \sin (3 e+4 f x)+450 c d \sin (5 e+4 f x)+720 c d \sin (4 e+5 f x)+750 d^2 \sin (e+2 f x)+750 d^2 \sin (3 e+2 f x)+1520 d^2 \sin (2 e+3 f x)+195 d^2 \sin (3 e+4 f x)+195 d^2 \sin (5 e+4 f x)+304 d^2 \sin (4 e+5 f x)\right )\right )}{15360 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 + Cos[e + f*x])^3*Sec[(e + f*x)/2]^6*Sec[e + f*x]^5*(240*(20*c^2 + 30*c*d + 13*d^2)*Cos[e + f*x]^5*(L
og[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) - Sec[e]*(80*(34*c^2 + 60*
c*d + 29*d^2)*Sin[f*x] - 240*(7*c^2 + 10*c*d + 3*d^2)*Sin[2*e + f*x] + 360*c^2*Sin[e + 2*f*x] + 1140*c*d*Sin[e
+ 2*f*x] + 750*d^2*Sin[e + 2*f*x] + 360*c^2*Sin[3*e + 2*f*x] + 1140*c*d*Sin[3*e + 2*f*x] + 750*d^2*Sin[3*e +
2*f*x] + 1840*c^2*Sin[2*e + 3*f*x] + 3360*c*d*Sin[2*e + 3*f*x] + 1520*d^2*Sin[2*e + 3*f*x] - 360*c^2*Sin[4*e +
3*f*x] - 240*c*d*Sin[4*e + 3*f*x] + 180*c^2*Sin[3*e + 4*f*x] + 450*c*d*Sin[3*e + 4*f*x] + 195*d^2*Sin[3*e + 4
*f*x] + 180*c^2*Sin[5*e + 4*f*x] + 450*c*d*Sin[5*e + 4*f*x] + 195*d^2*Sin[5*e + 4*f*x] + 440*c^2*Sin[4*e + 5*f
*x] + 720*c*d*Sin[4*e + 5*f*x] + 304*d^2*Sin[4*e + 5*f*x])))/(15360*f)

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Maple [A]  time = 0.06, size = 342, normalized size = 1.3 \begin{align*}{\frac{5\,{a}^{3}{c}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+6\,{\frac{{a}^{3}cd\tan \left ( fx+e \right ) }{f}}+{\frac{13\,{a}^{3}{d}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{13\,{a}^{3}{d}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+{\frac{11\,{a}^{3}{c}^{2}\tan \left ( fx+e \right ) }{3\,f}}+{\frac{15\,{a}^{3}cd\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{4\,f}}+{\frac{15\,{a}^{3}cd\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{4\,f}}+{\frac{38\,{a}^{3}{d}^{2}\tan \left ( fx+e \right ) }{15\,f}}+{\frac{19\,{a}^{3}{d}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15\,f}}+{\frac{3\,{a}^{3}{c}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+2\,{\frac{{a}^{3}cd\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{3\,{a}^{3}{d}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,f}}+{\frac{{a}^{3}{c}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}}+{\frac{{a}^{3}cd\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{2\,f}}+{\frac{{a}^{3}{d}^{2}\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))^3*(c+d*sec(f*x+e))^2,x)

[Out]

5/2/f*a^3*c^2*ln(sec(f*x+e)+tan(f*x+e))+6/f*a^3*c*d*tan(f*x+e)+13/8/f*a^3*d^2*sec(f*x+e)*tan(f*x+e)+13/8/f*a^3
*d^2*ln(sec(f*x+e)+tan(f*x+e))+11/3/f*a^3*c^2*tan(f*x+e)+15/4/f*a^3*c*d*sec(f*x+e)*tan(f*x+e)+15/4/f*a^3*c*d*l
n(sec(f*x+e)+tan(f*x+e))+38/15/f*a^3*d^2*tan(f*x+e)+19/15/f*a^3*d^2*tan(f*x+e)*sec(f*x+e)^2+3/2*a^3*c^2*sec(f*
x+e)*tan(f*x+e)/f+2/f*a^3*c*d*tan(f*x+e)*sec(f*x+e)^2+3/4/f*a^3*d^2*tan(f*x+e)*sec(f*x+e)^3+1/3/f*a^3*c^2*tan(
f*x+e)*sec(f*x+e)^2+1/2/f*a^3*c*d*tan(f*x+e)*sec(f*x+e)^3+1/5/f*a^3*d^2*tan(f*x+e)*sec(f*x+e)^4

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Maxima [A]  time = 1.01397, size = 620, normalized size = 2.41 \begin{align*} \frac{80 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c^{2} + 480 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c d + 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^{3} d^{2} + 240 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} d^{2} - 30 \, a^{3} c d{\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 )} - 45 \, a^{3} 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 )} - 180 \, a^{3} c^{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 )} - 360 \, a^{3} c 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 )} - 60 \, a^{3} 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^{3} c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 720 \, a^{3} c^{2} \tan \left (f x + e\right ) + 480 \, a^{3} c 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))^3*(c+d*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.516725, size = 587, normalized size = 2.28 \begin{align*} \frac{15 \,{\left (20 \, a^{3} c^{2} + 30 \, a^{3} c d + 13 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \,{\left (20 \, a^{3} c^{2} + 30 \, a^{3} c d + 13 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (24 \, a^{3} d^{2} + 8 \,{\left (55 \, a^{3} c^{2} + 90 \, a^{3} c d + 38 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \,{\left (12 \, a^{3} c^{2} + 30 \, a^{3} c d + 13 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \,{\left (5 \, a^{3} c^{2} + 30 \, a^{3} c d + 19 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} + 30 \,{\left (2 \, a^{3} c d + 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, 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))^3*(c+d*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

1/240*(15*(20*a^3*c^2 + 30*a^3*c*d + 13*a^3*d^2)*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 15*(20*a^3*c^2 + 30*a^
3*c*d + 13*a^3*d^2)*cos(f*x + e)^5*log(-sin(f*x + e) + 1) + 2*(24*a^3*d^2 + 8*(55*a^3*c^2 + 90*a^3*c*d + 38*a^
3*d^2)*cos(f*x + e)^4 + 15*(12*a^3*c^2 + 30*a^3*c*d + 13*a^3*d^2)*cos(f*x + e)^3 + 8*(5*a^3*c^2 + 30*a^3*c*d +
19*a^3*d^2)*cos(f*x + e)^2 + 30*(2*a^3*c*d + 3*a^3*d^2)*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^{3} \left (\int c^{2} \sec{\left (e + f x \right )}\, dx + \int 3 c^{2} \sec ^{2}{\left (e + f x \right )}\, dx + \int 3 c^{2} \sec ^{3}{\left (e + f x \right )}\, dx + \int c^{2} \sec ^{4}{\left (e + f x \right )}\, dx + \int d^{2} \sec ^{3}{\left (e + f x \right )}\, dx + \int 3 d^{2} \sec ^{4}{\left (e + f x \right )}\, dx + \int 3 d^{2} \sec ^{5}{\left (e + f x \right )}\, dx + \int d^{2} \sec ^{6}{\left (e + f x \right )}\, dx + \int 2 c d \sec ^{2}{\left (e + f x \right )}\, dx + \int 6 c d \sec ^{3}{\left (e + f x \right )}\, dx + \int 6 c d \sec ^{4}{\left (e + f x \right )}\, dx + \int 2 c d \sec ^{5}{\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))**3*(c+d*sec(f*x+e))**2,x)

[Out]

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

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Giac [A]  time = 1.28398, size = 532, normalized size = 2.07 \begin{align*} \frac{15 \,{\left (20 \, a^{3} c^{2} + 30 \, a^{3} c d + 13 \, a^{3} d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 15 \,{\left (20 \, a^{3} c^{2} + 30 \, a^{3} c d + 13 \, a^{3} d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (300 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} + 450 \, a^{3} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} + 195 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} - 1400 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 2100 \, a^{3} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 910 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 2560 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3840 \, a^{3} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 1664 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 2120 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 3660 \, a^{3} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 1330 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 660 \, a^{3} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1470 \, a^{3} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 765 \, a^{3} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1\right )}^{5}}}{120 \, 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))^3*(c+d*sec(f*x+e))^2,x, algorithm="giac")

[Out]

1/120*(15*(20*a^3*c^2 + 30*a^3*c*d + 13*a^3*d^2)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15*(20*a^3*c^2 + 30*a^3*
c*d + 13*a^3*d^2)*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(300*a^3*c^2*tan(1/2*f*x + 1/2*e)^9 + 450*a^3*c*d*tan
(1/2*f*x + 1/2*e)^9 + 195*a^3*d^2*tan(1/2*f*x + 1/2*e)^9 - 1400*a^3*c^2*tan(1/2*f*x + 1/2*e)^7 - 2100*a^3*c*d*
tan(1/2*f*x + 1/2*e)^7 - 910*a^3*d^2*tan(1/2*f*x + 1/2*e)^7 + 2560*a^3*c^2*tan(1/2*f*x + 1/2*e)^5 + 3840*a^3*c
*d*tan(1/2*f*x + 1/2*e)^5 + 1664*a^3*d^2*tan(1/2*f*x + 1/2*e)^5 - 2120*a^3*c^2*tan(1/2*f*x + 1/2*e)^3 - 3660*a
^3*c*d*tan(1/2*f*x + 1/2*e)^3 - 1330*a^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 660*a^3*c^2*tan(1/2*f*x + 1/2*e) + 1470*
a^3*c*d*tan(1/2*f*x + 1/2*e) + 765*a^3*d^2*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^5)/f