3.692 \(\int \frac{\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\)
Optimal. Leaf size=271 \[ \frac{\left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 b^4 (A+12 C)-15 a^4 b^2 C+6 a^6 C+2 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a \left (a^2 b^2 (A+6 C)-3 a^4 C+2 A b^4\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d} \]
[Out]
(-3*a*C*ArcTanh[Sin[c + d*x]])/(b^4*d) + ((2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[(Sqr
t[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) + ((A*b^2 + 3*a^2*C - 2*b^2*C)*Ta
n[c + d*x])/(2*b^3*(a^2 - b^2)*d) - ((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Se
c[c + d*x])^2) - (a*(2*A*b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Tan[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Sec[c
+ d*x]))
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Rubi [A] time = 1.02213, antiderivative size = 271, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used =
{4099, 4090, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (3 a^2 C+A b^2-2 b^2 C\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )}+\frac{\left (a^2 b^4 (A+12 C)-15 a^4 b^2 C+6 a^6 C+2 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a \left (a^2 b^2 (A+6 C)-3 a^4 C+2 A b^4\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac{3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]
[Out]
(-3*a*C*ArcTanh[Sin[c + d*x]])/(b^4*d) + ((2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A + 12*C))*ArcTanh[(Sqr
t[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) + ((A*b^2 + 3*a^2*C - 2*b^2*C)*Ta
n[c + d*x])/(2*b^3*(a^2 - b^2)*d) - ((A*b^2 + a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Se
c[c + d*x])^2) - (a*(2*A*b^4 - 3*a^4*C + a^2*b^2*(A + 6*C))*Tan[c + d*x])/(2*b^3*(a^2 - b^2)^2*d*(a + b*Sec[c
+ d*x]))
Rule 4099
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(d*(A*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f
*x])^(n - 1))/(b*f*(a^2 - b^2)*(m + 1)), x] + Dist[d/(b*(a^2 - b^2)*(m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)
*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) + a^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(m +
n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Rule 4090
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e
+ f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*(-(a*(b*B - a*C)) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Rule 4082
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Rule 3998
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rule 3831
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2659
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\sec ^2(c+d x) \left (2 \left (A b^2+a^2 C\right )-2 a b (A+C) \sec (c+d x)-\left (A b^2+3 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-b \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right )-a \left (a^2-b^2\right ) \left (A b^2-3 a^2 C+4 b^2 C\right ) \sec (c+d x)-b \left (a^2-b^2\right ) \left (A b^2+3 a^2 C-2 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-b^2 \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right )+6 a b \left (a^2-b^2\right )^2 C \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{(3 a C) \int \sec (c+d x) \, dx}{b^4}+\frac{\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b^5 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\left (2 A b^6+6 a^6 C-15 a^4 b^2 C+a^2 b^4 (A+12 C)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^5 \left (a^2-b^2\right )^2 d}\\ &=-\frac{3 a C \tanh ^{-1}(\sin (c+d x))}{b^4 d}+\frac{\left (a^2 A b^4+2 A b^6+6 a^6 C-15 a^4 b^2 C+12 a^2 b^4 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} b^4 (a+b)^{5/2} d}+\frac{\left (A b^2+3 a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}-\frac{\left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{a \left (2 A b^4-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 2.26337, size = 421, normalized size = 1.55 \[ \frac{\sec (c+d x) (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{a b^2 \left (a^2 C+A b^2\right ) \sin (c+d x)}{(a-b) (a+b)}+\frac{a b \left (-7 a^2 b^2 C+4 a^4 C-3 A b^4\right ) \sin (c+d x) (a \cos (c+d x)+b)}{(a-b)^2 (a+b)^2}-\frac{2 \left (a^2 b^4 (A+12 C)-15 a^4 b^2 C+6 a^6 C+2 A b^6\right ) (a \cos (c+d x)+b)^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{2 b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{2 b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+6 a C (a \cos (c+d x)+b)^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 a C (a \cos (c+d x)+b)^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{b^4 d (a+b \sec (c+d x))^3 (A \cos (2 (c+d x))+A+2 C)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]
[Out]
((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*((-2*(2*A*b^6 + 6*a^6*C - 15*a^4*b^2*C + a^2*b^4*(A
+ 12*C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x])^2)/(a^2 - b^2)^(5/2) + 6*a*
C*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*a*C*(b + a*Cos[c + d*x])^2*Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]] + (2*b*C*(b + a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]) + (2*b*C*(b + a*Cos[c + d*x])^2*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (a*b^2*(A*b^2
+ a^2*C)*Sin[c + d*x])/((a - b)*(a + b)) + (a*b*(-3*A*b^4 + 4*a^4*C - 7*a^2*b^2*C)*(b + a*Cos[c + d*x])*Sin[c
+ d*x])/((a - b)^2*(a + b)^2)))/(b^4*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)
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Maple [B] time = 0.098, size = 1167, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)
[Out]
1/d*a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A+4/d
*b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-4/d*a^
5/b^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+1/d*a
^4/b^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+8/d/
b/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a^3/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C+1/d/(
tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a^2/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*A-4/d*b/(tan
(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*A+4/d/b^3/(tan(1/
2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a^5/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C+1/d/b^2/(tan(1/2
*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a^4/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C-8/d/b/(tan(1/2*d*
x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2*a^3/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)*C+1/d/(a^4-2*a^2*b^2+b
^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*a^2+2/d*b^2/(a^4-2*a^2*b^2+b^4
)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+6/d/b^4/(a^4-2*a^2*b^2+b^4)/((a+
b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^6*C-15/d/b^2/(a^4-2*a^2*b^2+b^4)/((a+b
)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*a^4*C+12/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-
b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C*a^2-1/d*C/b^3/(tan(1/2*d*x+1/2*c)+1)-3/d*a*C
/b^4*ln(tan(1/2*d*x+1/2*c)+1)-1/d*C/b^3/(tan(1/2*d*x+1/2*c)-1)+3/d*a*C/b^4*ln(tan(1/2*d*x+1/2*c)-1)
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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(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 28.1077, size = 3416, normalized size = 12.61 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
[1/4*(((6*C*a^8 - 15*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6)*cos(d*x + c)^3 + 2*(6*C*a^7*b - 15*C*a^5*b^
3 + (A + 12*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c)^2 + (6*C*a^6*b^2 - 15*C*a^4*b^4 + (A + 12*C)*a^2*b^6 + 2*A*b^
8)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b
*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 6*((C*a^9 -
3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a^2*b^7)*co
s(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(sin(d*x + c) + 1) + 6*((C*a
^9 - 3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a^2*b^
7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2
*(2*C*a^6*b^3 - 6*C*a^4*b^5 + 6*C*a^2*b^7 - 2*C*b^9 + (6*C*a^8*b - 17*C*a^6*b^3 - (3*A - 13*C)*a^4*b^5 + (3*A
- 2*C)*a^2*b^7)*cos(d*x + c)^2 + (9*C*a^7*b^2 + (A - 25*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 4*(A - C)*a*b^8)*co
s(d*x + c))*sin(d*x + c))/((a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d*cos(d*x + c)^3 + 2*(a^7*b^5 - 3*a^5*
b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c)^2 + (a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c)), 1/2*((
(6*C*a^8 - 15*C*a^6*b^2 + (A + 12*C)*a^4*b^4 + 2*A*a^2*b^6)*cos(d*x + c)^3 + 2*(6*C*a^7*b - 15*C*a^5*b^3 + (A
+ 12*C)*a^3*b^5 + 2*A*a*b^7)*cos(d*x + c)^2 + (6*C*a^6*b^2 - 15*C*a^4*b^4 + (A + 12*C)*a^2*b^6 + 2*A*b^8)*cos(
d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - 3*((C*a
^9 - 3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a^2*b^
7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(sin(d*x + c) + 1) + 3*
((C*a^9 - 3*C*a^7*b^2 + 3*C*a^5*b^4 - C*a^3*b^6)*cos(d*x + c)^3 + 2*(C*a^8*b - 3*C*a^6*b^3 + 3*C*a^4*b^5 - C*a
^2*b^7)*cos(d*x + c)^2 + (C*a^7*b^2 - 3*C*a^5*b^4 + 3*C*a^3*b^6 - C*a*b^8)*cos(d*x + c))*log(-sin(d*x + c) + 1
) + (2*C*a^6*b^3 - 6*C*a^4*b^5 + 6*C*a^2*b^7 - 2*C*b^9 + (6*C*a^8*b - 17*C*a^6*b^3 - (3*A - 13*C)*a^4*b^5 + (3
*A - 2*C)*a^2*b^7)*cos(d*x + c)^2 + (9*C*a^7*b^2 + (A - 25*C)*a^5*b^4 - 5*(A - 4*C)*a^3*b^6 + 4*(A - C)*a*b^8)
*cos(d*x + c))*sin(d*x + c))/((a^8*b^4 - 3*a^6*b^6 + 3*a^4*b^8 - a^2*b^10)*d*cos(d*x + c)^3 + 2*(a^7*b^5 - 3*a
^5*b^7 + 3*a^3*b^9 - a*b^11)*d*cos(d*x + c)^2 + (a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*d*cos(d*x + c))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)
[Out]
Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**3, x)
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Giac [B] time = 1.31827, size = 703, normalized size = 2.59 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")
[Out]
((6*C*a^6 - 15*C*a^4*b^2 + A*a^2*b^4 + 12*C*a^2*b^4 + 2*A*b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*
b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4*b^4 - 2*a^2*b^6 + b^8)
*sqrt(-a^2 + b^2)) - 3*C*a*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + 3*C*a*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b
^4 - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 -
A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 +
4*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^6*tan(1/2*d*x + 1/2*c) - 5*C*a^5*b*tan(1/2*d*x + 1/2*c) + 7*C*a^4*b^2
*tan(1/2*d*x + 1/2*c) - A*a^3*b^3*tan(1/2*d*x + 1/2*c) + 8*C*a^3*b^3*tan(1/2*d*x + 1/2*c) + 3*A*a^2*b^4*tan(1/
2*d*x + 1/2*c) + 4*A*a*b^5*tan(1/2*d*x + 1/2*c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*d*x + 1/2*c)^2 - b*ta
n(1/2*d*x + 1/2*c)^2 - a - b)^2) - 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b^3))/d