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

Optimal. Leaf size=176 \[ -\frac{a^2 \left (-8 c^2 d+c^3-20 c d^2-8 d^3\right ) \tan (e+f x)}{6 d f}+\frac{a^2 \left (12 c^2+16 c d+7 d^2\right ) \tanh ^{-1}(\sin (e+f x))}{8 f}-\frac{a^2 \left (2 c (c-8 d)-21 d^2\right ) \tan (e+f x) \sec (e+f x)}{24 f}+\frac{a^2 \tan (e+f x) (c+d \sec (e+f x))^3}{4 d f}-\frac{a^2 (c-8 d) \tan (e+f x) (c+d \sec (e+f x))^2}{12 d f} \]

[Out]

(a^2*(12*c^2 + 16*c*d + 7*d^2)*ArcTanh[Sin[e + f*x]])/(8*f) - (a^2*(c^3 - 8*c^2*d - 20*c*d^2 - 8*d^3)*Tan[e +
f*x])/(6*d*f) - (a^2*(2*c*(c - 8*d) - 21*d^2)*Sec[e + f*x]*Tan[e + f*x])/(24*f) - (a^2*(c - 8*d)*(c + d*Sec[e
+ f*x])^2*Tan[e + f*x])/(12*d*f) + (a^2*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(4*d*f)

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Rubi [A]  time = 0.262041, antiderivative size = 234, normalized size of antiderivative = 1.33, 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, 90, 80, 50, 63, 217, 203} \[ \frac{a^2 \left (12 c^2+16 c d+7 d^2\right ) \tan (e+f x)}{8 f}+\frac{a^3 \left (12 c^2+16 c d+7 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{4 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{\left (12 c^2+16 c d+7 d^2\right ) \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{24 f}+\frac{d (5 c+2 d) \tan (e+f x) (a \sec (e+f x)+a)^2}{12 f}+\frac{d \tan (e+f x) (a \sec (e+f x)+a)^2 (c+d \sec (e+f x))}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*c^2 + 16*c*d + 7*d^2)*Tan[e + f*x])/(8*f) + (a^3*(12*c^2 + 16*c*d + 7*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]]) + (d*(
5*c + 2*d)*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(12*f) + ((12*c^2 + 16*c*d + 7*d^2)*(a^2 + a^2*Sec[e + f*x])*T
an[e + f*x])/(24*f) + (d*(a + a*Sec[e + f*x])^2*(c + d*Sec[e + f*x])*Tan[e + f*x])/(4*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))^2 (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)^{3/2} (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))^2 (c+d \sec (e+f x)) \tan (e+f x)}{4 f}+\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(a+a x)^{3/2} \left (-a^2 \left (4 c^2+2 c d+d^2\right )-a^2 d (5 c+2 d) x\right )}{\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{d (5 c+2 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x)) \tan (e+f x)}{4 f}-\frac{\left (a^2 \left (12 c^2+16 c d+7 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{d (5 c+2 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{\left (12 c^2+16 c d+7 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x)) \tan (e+f x)}{4 f}-\frac{\left (a^3 \left (12 c^2+16 c d+7 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^2 \left (12 c^2+16 c d+7 d^2\right ) \tan (e+f x)}{8 f}+\frac{d (5 c+2 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{\left (12 c^2+16 c d+7 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x)) \tan (e+f x)}{4 f}-\frac{\left (a^4 \left (12 c^2+16 c d+7 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^2 \left (12 c^2+16 c d+7 d^2\right ) \tan (e+f x)}{8 f}+\frac{d (5 c+2 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{\left (12 c^2+16 c d+7 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x)) \tan (e+f x)}{4 f}+\frac{\left (a^3 \left (12 c^2+16 c d+7 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^2 \left (12 c^2+16 c d+7 d^2\right ) \tan (e+f x)}{8 f}+\frac{d (5 c+2 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{\left (12 c^2+16 c d+7 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x)) \tan (e+f x)}{4 f}+\frac{\left (a^3 \left (12 c^2+16 c d+7 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^2 \left (12 c^2+16 c d+7 d^2\right ) \tan (e+f x)}{8 f}+\frac{a^3 \left (12 c^2+16 c d+7 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{d (5 c+2 d) (a+a \sec (e+f x))^2 \tan (e+f x)}{12 f}+\frac{\left (12 c^2+16 c d+7 d^2\right ) \left (a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{24 f}+\frac{d (a+a \sec (e+f x))^2 (c+d \sec (e+f x)) \tan (e+f x)}{4 f}\\ \end{align*}

Mathematica [B]  time = 1.00136, size = 479, normalized size = 2.72 \[ -\frac{a^2 \sec ^4(e+f x) \left (12 \left (12 c^2+16 c d+7 d^2\right ) \cos (2 (e+f x)) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )+3 \left (12 c^2+16 c d+7 d^2\right ) \cos (4 (e+f x)) \left (\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-24 c^2 \sin (e+f x)-96 c^2 \sin (2 (e+f x))-24 c^2 \sin (3 (e+f x))-48 c^2 \sin (4 (e+f x))+108 c^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-108 c^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-96 c d \sin (e+f x)-224 c d \sin (2 (e+f x))-96 c d \sin (3 (e+f x))-80 c d \sin (4 (e+f x))+144 c d \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-144 c d \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-90 d^2 \sin (e+f x)-128 d^2 \sin (2 (e+f x))-42 d^2 \sin (3 (e+f x))-32 d^2 \sin (4 (e+f x))+63 d^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-63 d^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{192 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Sec[e + f*x]^4*(108*c^2*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 144*c*d*Log[Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2]] + 63*d^2*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 12*(12*c^2 + 16*c*d + 7*d^2)*Cos[2*(e + f*x)]*
(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + 3*(12*c^2 + 16*c*d + 7
*d^2)*Cos[4*(e + f*x)]*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) -
108*c^2*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 144*c*d*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 63*d^2*
Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] - 24*c^2*Sin[e + f*x] - 96*c*d*Sin[e + f*x] - 90*d^2*Sin[e + f*x] - 9
6*c^2*Sin[2*(e + f*x)] - 224*c*d*Sin[2*(e + f*x)] - 128*d^2*Sin[2*(e + f*x)] - 24*c^2*Sin[3*(e + f*x)] - 96*c*
d*Sin[3*(e + f*x)] - 42*d^2*Sin[3*(e + f*x)] - 48*c^2*Sin[4*(e + f*x)] - 80*c*d*Sin[4*(e + f*x)] - 32*d^2*Sin[
4*(e + f*x)]))/(192*f)

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Maple [A]  time = 0.052, size = 268, normalized size = 1.5 \begin{align*}{\frac{3\,{a}^{2}{c}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{2\,f}}+{\frac{10\,{a}^{2}cd\tan \left ( fx+e \right ) }{3\,f}}+{\frac{7\,{a}^{2}{d}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{8\,f}}+{\frac{7\,{a}^{2}{d}^{2}\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{8\,f}}+2\,{\frac{{a}^{2}{c}^{2}\tan \left ( fx+e \right ) }{f}}+2\,{\frac{{a}^{2}cd\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{f}}+2\,{\frac{{a}^{2}cd\ln \left ( \sec \left ( fx+e \right ) +\tan \left ( fx+e \right ) \right ) }{f}}+{\frac{4\,{a}^{2}{d}^{2}\tan \left ( fx+e \right ) }{3\,f}}+{\frac{2\,{a}^{2}{d}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}}+{\frac{{a}^{2}{c}^{2}\sec \left ( fx+e \right ) \tan \left ( fx+e \right ) }{2\,f}}+{\frac{2\,{a}^{2}cd\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3\,f}}+{\frac{{a}^{2}{d}^{2}\tan \left ( fx+e \right ) \left ( \sec \left ( fx+e \right ) \right ) ^{3}}{4\,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))^2,x)

[Out]

3/2/f*a^2*c^2*ln(sec(f*x+e)+tan(f*x+e))+10/3/f*a^2*c*d*tan(f*x+e)+7/8/f*a^2*d^2*sec(f*x+e)*tan(f*x+e)+7/8/f*a^
2*d^2*ln(sec(f*x+e)+tan(f*x+e))+2/f*a^2*c^2*tan(f*x+e)+2/f*a^2*c*d*sec(f*x+e)*tan(f*x+e)+2/f*a^2*c*d*ln(sec(f*
x+e)+tan(f*x+e))+4/3/f*a^2*d^2*tan(f*x+e)+2/3/f*a^2*d^2*tan(f*x+e)*sec(f*x+e)^2+1/2*a^2*c^2*sec(f*x+e)*tan(f*x
+e)/f+2/3/f*a^2*c*d*tan(f*x+e)*sec(f*x+e)^2+1/4/f*a^2*d^2*tan(f*x+e)*sec(f*x+e)^3

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Maxima [A]  time = 1.01973, size = 437, normalized size = 2.48 \begin{align*} \frac{32 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} c d + 32 \,{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{2} d^{2} - 3 \, a^{2} 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 )} - 12 \, a^{2} 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 )} - 48 \, a^{2} 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 )} - 12 \, a^{2} 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 )} + 48 \, a^{2} c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + 96 \, a^{2} c^{2} \tan \left (f x + e\right ) + 96 \, a^{2} c d \tan \left (f x + e\right )}{48 \, 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))^2,x, algorithm="maxima")

[Out]

1/48*(32*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*c*d + 32*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^2*d^2 - 3*a^2*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*l
og(sin(f*x + e) - 1)) - 12*a^2*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)) - 48*a^2*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))
- 12*a^2*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)) + 48*a^2*c^
2*log(sec(f*x + e) + tan(f*x + e)) + 96*a^2*c^2*tan(f*x + e) + 96*a^2*c*d*tan(f*x + e))/f

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Fricas [A]  time = 0.510715, size = 490, normalized size = 2.78 \begin{align*} \frac{3 \,{\left (12 \, a^{2} c^{2} + 16 \, a^{2} c d + 7 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \,{\left (12 \, a^{2} c^{2} + 16 \, a^{2} c d + 7 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left (6 \, a^{2} d^{2} + 16 \,{\left (3 \, a^{2} c^{2} + 5 \, a^{2} c d + 2 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (4 \, a^{2} c^{2} + 16 \, a^{2} c d + 7 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 16 \,{\left (a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \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))^2,x, algorithm="fricas")

[Out]

1/48*(3*(12*a^2*c^2 + 16*a^2*c*d + 7*a^2*d^2)*cos(f*x + e)^4*log(sin(f*x + e) + 1) - 3*(12*a^2*c^2 + 16*a^2*c*
d + 7*a^2*d^2)*cos(f*x + e)^4*log(-sin(f*x + e) + 1) + 2*(6*a^2*d^2 + 16*(3*a^2*c^2 + 5*a^2*c*d + 2*a^2*d^2)*c
os(f*x + e)^3 + 3*(4*a^2*c^2 + 16*a^2*c*d + 7*a^2*d^2)*cos(f*x + e)^2 + 16*(a^2*c*d + a^2*d^2)*cos(f*x + e))*s
in(f*x + e))/(f*cos(f*x + e)^4)

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

[Out]

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

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Giac [B]  time = 1.40344, size = 452, normalized size = 2.57 \begin{align*} \frac{3 \,{\left (12 \, a^{2} c^{2} + 16 \, a^{2} c d + 7 \, a^{2} d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right ) - 3 \,{\left (12 \, a^{2} c^{2} + 16 \, a^{2} c d + 7 \, a^{2} d^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right ) - \frac{2 \,{\left (36 \, a^{2} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 48 \, a^{2} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 21 \, a^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 132 \, a^{2} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 176 \, a^{2} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 77 \, a^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 156 \, a^{2} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 272 \, a^{2} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 83 \, a^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 60 \, a^{2} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 144 \, a^{2} c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 75 \, a^{2} 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 )}^{4}}}{24 \, 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))^2,x, algorithm="giac")

[Out]

1/24*(3*(12*a^2*c^2 + 16*a^2*c*d + 7*a^2*d^2)*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 3*(12*a^2*c^2 + 16*a^2*c*d
+ 7*a^2*d^2)*log(abs(tan(1/2*f*x + 1/2*e) - 1)) - 2*(36*a^2*c^2*tan(1/2*f*x + 1/2*e)^7 + 48*a^2*c*d*tan(1/2*f*
x + 1/2*e)^7 + 21*a^2*d^2*tan(1/2*f*x + 1/2*e)^7 - 132*a^2*c^2*tan(1/2*f*x + 1/2*e)^5 - 176*a^2*c*d*tan(1/2*f*
x + 1/2*e)^5 - 77*a^2*d^2*tan(1/2*f*x + 1/2*e)^5 + 156*a^2*c^2*tan(1/2*f*x + 1/2*e)^3 + 272*a^2*c*d*tan(1/2*f*
x + 1/2*e)^3 + 83*a^2*d^2*tan(1/2*f*x + 1/2*e)^3 - 60*a^2*c^2*tan(1/2*f*x + 1/2*e) - 144*a^2*c*d*tan(1/2*f*x +
1/2*e) - 75*a^2*d^2*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^4)/f