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

Optimal. Leaf size=188 \[ -\frac{a^3 \sqrt{c-d} \left (2 c^2+6 c d+7 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c-d} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{c+d}}\right )}{d^3 f (c+d)^{5/2}}-\frac{a^3 (c-d) (2 c+5 d) \tan (e+f x)}{2 d^2 f (c+d)^2 (c+d \sec (e+f x))}-\frac{(c-d) \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{2 d f (c+d) (c+d \sec (e+f x))^2}+\frac{a^3 \tanh ^{-1}(\sin (e+f x))}{d^3 f} \]

[Out]

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

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Rubi [A]  time = 0.394444, antiderivative size = 301, normalized size of antiderivative = 1.6, number of steps used = 9, number of rules used = 9, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.29, Rules used = {3987, 98, 149, 157, 63, 217, 203, 93, 205} \[ \frac{a^4 \sqrt{c-d} \left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{c+d} \sqrt{a \sec (e+f x)+a}}{\sqrt{c-d} \sqrt{a-a \sec (e+f x)}}\right )}{d^3 f (c+d)^{5/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{a^3 (c-d) (2 c+5 d) \tan (e+f x)}{2 d^2 f (c+d)^2 (c+d \sec (e+f x))}-\frac{(c-d) \tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right )}{2 d f (c+d) (c+d \sec (e+f x))^2}+\frac{2 a^4 \tan (e+f x) \tan ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a (\sec (e+f x)+1)}}\right )}{d^3 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*ArcTan[Sqrt[a - a*Sec[e + f*x]]/Sqrt[a*(1 + Sec[e + f*x])]]*Tan[e + f*x])/(d^3*f*Sqrt[a - a*Sec[e + f*x
]]*Sqrt[a + a*Sec[e + f*x]]) + (a^4*Sqrt[c - d]*(2*c^2 + 6*c*d + 7*d^2)*ArcTan[(Sqrt[c + d]*Sqrt[a + a*Sec[e +
f*x]])/(Sqrt[c - d]*Sqrt[a - a*Sec[e + f*x]])]*Tan[e + f*x])/(d^3*(c + d)^(5/2)*f*Sqrt[a - a*Sec[e + f*x]]*Sq
rt[a + a*Sec[e + f*x]]) - ((c - d)*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(2*d*(c + d)*f*(c + d*Sec[e + f*x])^
2) - (a^3*(c - d)*(2*c + 5*d)*Tan[e + f*x])/(2*d^2*(c + d)^2*f*(c + d*Sec[e + f*x]))

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 149

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] && IntegerQ[m]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, 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])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 205

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

Rubi steps

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

Mathematica [C]  time = 3.37798, size = 393, normalized size = 2.09 \[ \frac{a^3 \sec ^6\left (\frac{1}{2} (e+f x)\right ) (\sec (e+f x)+1)^3 (c \cos (e+f x)+d) \left (\frac{d (c-d) \sec (e) \left (\left (5 c^2 d^2+6 c^3 d+2 c^4+12 c d^3+2 d^4\right ) \sin (e)-c \left (-d \left (c^2+6 c d+2 d^2\right ) \sin (2 e+f x)+c \left (2 c^2+6 c d+d^2\right ) \sin (e+2 f x)+d \left (7 c^2+18 c d+2 d^2\right ) \sin (f x)\right )\right )}{c^2 (c+d)^2}+\frac{4 \left (4 c^2 d+2 c^3+c d^2-7 d^3\right ) (\sin (e)+i \cos (e)) (c \cos (e+f x)+d)^2 \tan ^{-1}\left (\frac{(\sin (e)+i \cos (e)) \left (\tan \left (\frac{f x}{2}\right ) (c \cos (e)-d)+c \sin (e)\right )}{\sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}\right )}{(c+d)^2 \sqrt{c^2-d^2} \sqrt{(\cos (e)-i \sin (e))^2}}-4 (c \cos (e+f x)+d)^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+4 (c \cos (e+f x)+d)^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{32 d^3 f (c+d \sec (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(d + c*Cos[e + f*x])*Sec[(e + f*x)/2]^6*(1 + Sec[e + f*x])^3*(-4*(d + c*Cos[e + f*x])^2*Log[Cos[(e + f*x)
/2] - Sin[(e + f*x)/2]] + 4*(d + c*Cos[e + f*x])^2*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (4*(2*c^3 + 4*c^
2*d + c*d^2 - 7*d^3)*ArcTan[((I*Cos[e] + Sin[e])*(c*Sin[e] + (-d + c*Cos[e])*Tan[(f*x)/2]))/(Sqrt[c^2 - d^2]*S
qrt[(Cos[e] - I*Sin[e])^2])]*(d + c*Cos[e + f*x])^2*(I*Cos[e] + Sin[e]))/((c + d)^2*Sqrt[c^2 - d^2]*Sqrt[(Cos[
e] - I*Sin[e])^2]) + ((c - d)*d*Sec[e]*((2*c^4 + 6*c^3*d + 5*c^2*d^2 + 12*c*d^3 + 2*d^4)*Sin[e] - c*(d*(7*c^2
+ 18*c*d + 2*d^2)*Sin[f*x] - d*(c^2 + 6*c*d + 2*d^2)*Sin[2*e + f*x] + c*(2*c^2 + 6*c*d + d^2)*Sin[e + 2*f*x]))
)/(c^2*(c + d)^2)))/(32*d^3*f*(c + d*Sec[e + f*x])^3)

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Maple [B]  time = 0.127, size = 768, normalized size = 4.1 \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)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^3,x)

[Out]

1/f*a^3/d^3*ln(tan(1/2*f*x+1/2*e)+1)-1/f*a^3/d^3*ln(tan(1/2*f*x+1/2*e)-1)+2/f*a^3/d^2/(tan(1/2*f*x+1/2*e)^2*c-
tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*c^3+1/f*a^3/d/(tan(1/2*f*x+1/2*e)^2*c-tan(1
/2*f*x+1/2*e)^2*d-c-d)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3*c^2-8/f*a^3*c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*
x+1/2*e)^2*d-c-d)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3-2/f*a^3/d^2/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e
)^2*d-c-d)^2/(c+d)*tan(1/2*f*x+1/2*e)*c^2-5/f*a^3/d*c/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c
+d)*tan(1/2*f*x+1/2*e)-2/f*a^3/d^3/(c^2+2*c*d+d^2)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)
*(c-d))^(1/2))*c^3-4/f*a^3/d^2/(c^2+2*c*d+d^2)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-
d))^(1/2))*c^2-1/f*a^3/d*c/(c^2+2*c*d+d^2)/((c+d)*(c-d))^(1/2)*arctanh(tan(1/2*f*x+1/2*e)*(c-d)/((c+d)*(c-d))^
(1/2))+5/f*a^3*d/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3+7/
f*a^3/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/2*e)^2*d-c-d)^2/(c+d)*tan(1/2*f*x+1/2*e)+7/f*a^3/(c^2+2*c*d+d^2)/(
(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+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.68489, size = 2534, normalized size = 13.48 \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)*(a+a*sec(f*x+e))^3/(c+d*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/4*((2*a^3*c^2*d^2 + 6*a^3*c*d^3 + 7*a^3*d^4 + (2*a^3*c^4 + 6*a^3*c^3*d + 7*a^3*c^2*d^2)*cos(f*x + e)^2 + 2*
(2*a^3*c^3*d + 6*a^3*c^2*d^2 + 7*a^3*c*d^3)*cos(f*x + e))*sqrt((c - d)/(c + d))*log((2*c*d*cos(f*x + e) - (c^2
- 2*d^2)*cos(f*x + e)^2 - 2*(c^2 + c*d + (c*d + d^2)*cos(f*x + e))*sqrt((c - d)/(c + d))*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*(a^3*c^2*d^2 + 2*a^3*c*d^3 + a^3*d^4 + (a^3*c^4 +
2*a^3*c^3*d + a^3*c^2*d^2)*cos(f*x + e)^2 + 2*(a^3*c^3*d + 2*a^3*c^2*d^2 + a^3*c*d^3)*cos(f*x + e))*log(sin(f
*x + e) + 1) - 2*(a^3*c^2*d^2 + 2*a^3*c*d^3 + a^3*d^4 + (a^3*c^4 + 2*a^3*c^3*d + a^3*c^2*d^2)*cos(f*x + e)^2 +
2*(a^3*c^3*d + 2*a^3*c^2*d^2 + a^3*c*d^3)*cos(f*x + e))*log(-sin(f*x + e) + 1) - 2*(3*a^3*c^2*d^2 + 3*a^3*c*d
^3 - 6*a^3*d^4 + (2*a^3*c^3*d + 4*a^3*c^2*d^2 - 5*a^3*c*d^3 - a^3*d^4)*cos(f*x + e))*sin(f*x + e))/((c^4*d^3 +
2*c^3*d^4 + c^2*d^5)*f*cos(f*x + e)^2 + 2*(c^3*d^4 + 2*c^2*d^5 + c*d^6)*f*cos(f*x + e) + (c^2*d^5 + 2*c*d^6 +
d^7)*f), -1/2*((2*a^3*c^2*d^2 + 6*a^3*c*d^3 + 7*a^3*d^4 + (2*a^3*c^4 + 6*a^3*c^3*d + 7*a^3*c^2*d^2)*cos(f*x +
e)^2 + 2*(2*a^3*c^3*d + 6*a^3*c^2*d^2 + 7*a^3*c*d^3)*cos(f*x + e))*sqrt(-(c - d)/(c + d))*arctan(-(d*cos(f*x
+ e) + c)*sqrt(-(c - d)/(c + d))/((c - d)*sin(f*x + e))) - (a^3*c^2*d^2 + 2*a^3*c*d^3 + a^3*d^4 + (a^3*c^4 + 2
*a^3*c^3*d + a^3*c^2*d^2)*cos(f*x + e)^2 + 2*(a^3*c^3*d + 2*a^3*c^2*d^2 + a^3*c*d^3)*cos(f*x + e))*log(sin(f*x
+ e) + 1) + (a^3*c^2*d^2 + 2*a^3*c*d^3 + a^3*d^4 + (a^3*c^4 + 2*a^3*c^3*d + a^3*c^2*d^2)*cos(f*x + e)^2 + 2*(
a^3*c^3*d + 2*a^3*c^2*d^2 + a^3*c*d^3)*cos(f*x + e))*log(-sin(f*x + e) + 1) + (3*a^3*c^2*d^2 + 3*a^3*c*d^3 - 6
*a^3*d^4 + (2*a^3*c^3*d + 4*a^3*c^2*d^2 - 5*a^3*c*d^3 - a^3*d^4)*cos(f*x + e))*sin(f*x + e))/((c^4*d^3 + 2*c^3
*d^4 + c^2*d^5)*f*cos(f*x + e)^2 + 2*(c^3*d^4 + 2*c^2*d^5 + c*d^6)*f*cos(f*x + e) + (c^2*d^5 + 2*c*d^6 + d^7)*
f)]

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

[Out]

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

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Giac [B]  time = 1.36028, size = 528, normalized size = 2.81 \begin{align*} \frac{\frac{a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{d^{3}} - \frac{a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{d^{3}} + \frac{{\left (2 \, a^{3} c^{3} + 4 \, a^{3} c^{2} d + a^{3} c d^{2} - 7 \, a^{3} d^{3}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-c^{2} + d^{2}}}\right )\right )}}{{\left (c^{2} d^{3} + 2 \, c d^{4} + d^{5}\right )} \sqrt{-c^{2} + d^{2}}} + \frac{2 \, a^{3} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + a^{3} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 8 \, a^{3} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 5 \, a^{3} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, a^{3} c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 7 \, a^{3} c^{2} d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, a^{3} c d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, a^{3} d^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )}{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c - d\right )}^{2}}}{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))^3,x, algorithm="giac")

[Out]

(a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/d^3 - a^3*log(abs(tan(1/2*f*x + 1/2*e) - 1))/d^3 + (2*a^3*c^3 + 4*a^3*
c^2*d + a^3*c*d^2 - 7*a^3*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^2*d^3 + 2*c*d^4 + d^5)*sqrt(-c^2 + d^2)) + (2*a^3*c^3*tan(
1/2*f*x + 1/2*e)^3 + a^3*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 8*a^3*c*d^2*tan(1/2*f*x + 1/2*e)^3 + 5*a^3*d^3*tan(1/2
*f*x + 1/2*e)^3 - 2*a^3*c^3*tan(1/2*f*x + 1/2*e) - 7*a^3*c^2*d*tan(1/2*f*x + 1/2*e) + 2*a^3*c*d^2*tan(1/2*f*x
+ 1/2*e) + 7*a^3*d^3*tan(1/2*f*x + 1/2*e))/((c^2*d^2 + 2*c*d^3 + d^4)*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*
x + 1/2*e)^2 - c - d)^2))/f