3.808 \(\int \frac{\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\)
Optimal. Leaf size=289 \[ -\frac{\left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )}-\frac{a \left (-5 a^2 b^3 B+15 a^3 b^2 C+2 a^4 b B-6 a^5 C-12 a b^4 C+6 b^5 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a^2 \left (a^2 b B-3 a^3 C+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d} \]
[Out]
((b*B - 3*a*C)*ArcTanh[Sin[c + d*x]])/(b^4*d) - (a*(2*a^4*b*B - 5*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 15*a^3*b^2*C
- 12*a*b^4*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) - ((a*
b*B - 3*a^2*C + 2*b^2*C)*Tan[c + d*x])/(2*b^3*(a^2 - b^2)*d) + (a*(b*B - a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*
b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - (a^2*(a^2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*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 = 1.42482, antiderivative size = 289, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 9, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.225, Rules used =
{4072, 4029, 4090, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{\left (-3 a^2 C+a b B+2 b^2 C\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )}-\frac{a \left (-5 a^2 b^3 B+15 a^3 b^2 C+2 a^4 b B-6 a^5 C-12 a b^4 C+6 b^5 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d (a-b)^{5/2} (a+b)^{5/2}}+\frac{a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{a^2 \left (a^2 b B-3 a^3 C+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{2 b^3 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]
[Out]
((b*B - 3*a*C)*ArcTanh[Sin[c + d*x]])/(b^4*d) - (a*(2*a^4*b*B - 5*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 15*a^3*b^2*C
- 12*a*b^4*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*b^4*(a + b)^(5/2)*d) - ((a*
b*B - 3*a^2*C + 2*b^2*C)*Tan[c + d*x])/(2*b^3*(a^2 - b^2)*d) + (a*(b*B - a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*
b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - (a^2*(a^2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*C)*Tan[c + d*x])/(2*b^3*
(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))
Rule 4072
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rule 4029
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*d^2*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])
^(n - 2))/(b*f*(m + 1)*(a^2 - b^2)), x] - Dist[d/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*
Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*(n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) -
d*B*(a^2*(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a
*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 1]
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 (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=\int \frac{\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx\\ &=\frac{a (b B-a C) \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 a (b B-a C)-2 b (b B-a C) \sec (c+d x)-\left (a b B-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{a (b B-a C) \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^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 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 (a b \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right )+\left (a^2-b^2\right ) \left (a^2 b B-2 b^3 B-3 a^3 C+4 a b^2 C\right ) \sec (c+d x)-b \left (a^2-b^2\right ) \left (a b B-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 B-3 a^2 C+2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (b B-a C) \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^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 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 (a b^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right )+2 b \left (a^2-b^2\right )^2 (b B-3 a 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 B-3 a^2 C+2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (b B-a C) \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^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{(b B-3 a C) \int \sec (c+d x) \, dx}{b^4}-\frac{\left (a \left (2 a^4 b B-5 a^2 b^3 B+6 b^5 B-6 a^5 C+15 a^3 b^2 C-12 a b^4 C\right )\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{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{\left (a b B-3 a^2 C+2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (b B-a C) \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^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 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 (a \left (2 a^4 b B-5 a^2 b^3 B+6 b^5 B-6 a^5 C+15 a^3 b^2 C-12 a b^4 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 b^5 \left (a^2-b^2\right )^2}\\ &=\frac{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{\left (a b B-3 a^2 C+2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (b B-a C) \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^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 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 (a \left (2 a^4 b B-5 a^2 b^3 B+6 b^5 B-6 a^5 C+15 a^3 b^2 C-12 a b^4 C\right )\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{(b B-3 a C) \tanh ^{-1}(\sin (c+d x))}{b^4 d}-\frac{a \left (2 a^4 b B-5 a^2 b^3 B+6 b^5 B-6 a^5 C+15 a^3 b^2 C-12 a 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 B-3 a^2 C+2 b^2 C\right ) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{a (b B-a C) \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^2 \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 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 = 6.44998, size = 418, normalized size = 1.45 \[ \frac{a^2 b B \sin (c+d x)-a^3 C \sin (c+d x)}{2 b^2 d (b-a) (a+b) (a \cos (c+d x)+b)^2}+\frac{5 a^2 b^3 B \sin (c+d x)-7 a^3 b^2 C \sin (c+d x)-2 a^4 b B \sin (c+d x)+4 a^5 C \sin (c+d x)}{2 b^3 d (b-a)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac{a \left (-5 a^2 b^3 B+15 a^3 b^2 C+2 a^4 b B-6 a^5 C-12 a b^4 C+6 b^5 B\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^4 d \sqrt{a^2-b^2} \left (b^2-a^2\right )^2}+\frac{(3 a C-b B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{b^4 d}+\frac{(b B-3 a C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{b^4 d}+\frac{C \sin \left (\frac{1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{C \sin \left (\frac{1}{2} (c+d x)\right )}{b^3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sec[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]
[Out]
(a*(2*a^4*b*B - 5*a^2*b^3*B + 6*b^5*B - 6*a^5*C + 15*a^3*b^2*C - 12*a*b^4*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2
])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b^2]*(-a^2 + b^2)^2*d) + ((-(b*B) + 3*a*C)*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]])/(b^4*d) + ((b*B - 3*a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(b^4*d) + (C*Sin[(c + d*x)/2])/
(b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (C*Sin[(c + d*x)/2])/(b^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/
2])) + (a^2*b*B*Sin[c + d*x] - a^3*C*Sin[c + d*x])/(2*b^2*(-a + b)*(a + b)*d*(b + a*Cos[c + d*x])^2) + (-2*a^4
*b*B*Sin[c + d*x] + 5*a^2*b^3*B*Sin[c + d*x] + 4*a^5*C*Sin[c + d*x] - 7*a^3*b^2*C*Sin[c + d*x])/(2*b^3*(-a + b
)^2*(a + b)^2*d*(b + a*Cos[c + d*x]))
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Maple [B] time = 0.101, size = 1406, normalized size = 4.9 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)
[Out]
2/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*B
-1/d*a^3/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-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B-
6/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*B-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*a^3/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-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C-2/
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-b)^2*tan(1/2*d*x+1/2*c)*B-1/d*a^3/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+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B+6/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-b)^2*tan(1/2*d*x+1/2*c)*B+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-b)^2*tan(1/2*d*x+1/2*c)*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-b)^2*tan(1/2*d*x+1/2*c)*C-8/d*a^3/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+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*C-2/d*a^5/b^3/(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))*B+5/d*a^3/b/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*ta
n(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-6/d*a*b/(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))*B+6/d*a^6/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))*C-15/d*a^4/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))*C+12/d*a^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))*C-1/d*C/b^3/(tan(1/2*d*x+1/2*c)+1)+1/d/b^3*ln(tan(1/2*d*x+1/2*c)+1)*B-3/d/b^4*ln(tan(1
/2*d*x+1/2*c)+1)*a*C-1/d*C/b^3/(tan(1/2*d*x+1/2*c)-1)-1/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*B+3/d/b^4*ln(tan(1/2*d*
x+1/2*c)-1)*a*C
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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*(B*sec(d*x+c)+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 = 114.551, size = 4591, normalized size = 15.89 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
[-1/4*(((6*C*a^8 - 2*B*a^7*b - 15*C*a^6*b^2 + 5*B*a^5*b^3 + 12*C*a^4*b^4 - 6*B*a^3*b^5)*cos(d*x + c)^3 + 2*(6*
C*a^7*b - 2*B*a^6*b^2 - 15*C*a^5*b^3 + 5*B*a^4*b^4 + 12*C*a^3*b^5 - 6*B*a^2*b^6)*cos(d*x + c)^2 + (6*C*a^6*b^2
- 2*B*a^5*b^3 - 15*C*a^4*b^4 + 5*B*a^3*b^5 + 12*C*a^2*b^6 - 6*B*a*b^7)*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)) + 2*((3*C*a^9 - B*a^8*b - 9*C*a^7*b^2 + 3*B*a^6*b^3 + 9
*C*a^5*b^4 - 3*B*a^4*b^5 - 3*C*a^3*b^6 + B*a^2*b^7)*cos(d*x + c)^3 + 2*(3*C*a^8*b - B*a^7*b^2 - 9*C*a^6*b^3 +
3*B*a^5*b^4 + 9*C*a^4*b^5 - 3*B*a^3*b^6 - 3*C*a^2*b^7 + B*a*b^8)*cos(d*x + c)^2 + (3*C*a^7*b^2 - B*a^6*b^3 - 9
*C*a^5*b^4 + 3*B*a^4*b^5 + 9*C*a^3*b^6 - 3*B*a^2*b^7 - 3*C*a*b^8 + B*b^9)*cos(d*x + c))*log(sin(d*x + c) + 1)
- 2*((3*C*a^9 - B*a^8*b - 9*C*a^7*b^2 + 3*B*a^6*b^3 + 9*C*a^5*b^4 - 3*B*a^4*b^5 - 3*C*a^3*b^6 + B*a^2*b^7)*cos
(d*x + c)^3 + 2*(3*C*a^8*b - B*a^7*b^2 - 9*C*a^6*b^3 + 3*B*a^5*b^4 + 9*C*a^4*b^5 - 3*B*a^3*b^6 - 3*C*a^2*b^7 +
B*a*b^8)*cos(d*x + c)^2 + (3*C*a^7*b^2 - B*a^6*b^3 - 9*C*a^5*b^4 + 3*B*a^4*b^5 + 9*C*a^3*b^6 - 3*B*a^2*b^7 -
3*C*a*b^8 + B*b^9)*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 - 2*B*a^7*b^2 - 17*C*a^6*b^3 + 7*B*a^5*b^4 + 13*C*a^4*b^5 - 5*B*a^3*b^6 - 2*C*a^2*b^7)*cos(d*x +
c)^2 + (9*C*a^7*b^2 - 3*B*a^6*b^3 - 25*C*a^5*b^4 + 9*B*a^4*b^5 + 20*C*a^3*b^6 - 6*B*a^2*b^7 - 4*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)), 1/2*(((6
*C*a^8 - 2*B*a^7*b - 15*C*a^6*b^2 + 5*B*a^5*b^3 + 12*C*a^4*b^4 - 6*B*a^3*b^5)*cos(d*x + c)^3 + 2*(6*C*a^7*b -
2*B*a^6*b^2 - 15*C*a^5*b^3 + 5*B*a^4*b^4 + 12*C*a^3*b^5 - 6*B*a^2*b^6)*cos(d*x + c)^2 + (6*C*a^6*b^2 - 2*B*a^5
*b^3 - 15*C*a^4*b^4 + 5*B*a^3*b^5 + 12*C*a^2*b^6 - 6*B*a*b^7)*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 - B*a^8*b - 9*C*a^7*b^2 + 3*B*a^6*b^3 + 9
*C*a^5*b^4 - 3*B*a^4*b^5 - 3*C*a^3*b^6 + B*a^2*b^7)*cos(d*x + c)^3 + 2*(3*C*a^8*b - B*a^7*b^2 - 9*C*a^6*b^3 +
3*B*a^5*b^4 + 9*C*a^4*b^5 - 3*B*a^3*b^6 - 3*C*a^2*b^7 + B*a*b^8)*cos(d*x + c)^2 + (3*C*a^7*b^2 - B*a^6*b^3 - 9
*C*a^5*b^4 + 3*B*a^4*b^5 + 9*C*a^3*b^6 - 3*B*a^2*b^7 - 3*C*a*b^8 + B*b^9)*cos(d*x + c))*log(sin(d*x + c) + 1)
+ ((3*C*a^9 - B*a^8*b - 9*C*a^7*b^2 + 3*B*a^6*b^3 + 9*C*a^5*b^4 - 3*B*a^4*b^5 - 3*C*a^3*b^6 + B*a^2*b^7)*cos(d
*x + c)^3 + 2*(3*C*a^8*b - B*a^7*b^2 - 9*C*a^6*b^3 + 3*B*a^5*b^4 + 9*C*a^4*b^5 - 3*B*a^3*b^6 - 3*C*a^2*b^7 + B
*a*b^8)*cos(d*x + c)^2 + (3*C*a^7*b^2 - B*a^6*b^3 - 9*C*a^5*b^4 + 3*B*a^4*b^5 + 9*C*a^3*b^6 - 3*B*a^2*b^7 - 3*
C*a*b^8 + B*b^9)*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 - 2*B*a^7*b^2 - 17*C*a^6*b^3 + 7*B*a^5*b^4 + 13*C*a^4*b^5 - 5*B*a^3*b^6 - 2*C*a^2*b^7)*cos(d*x + c)^
2 + (9*C*a^7*b^2 - 3*B*a^6*b^3 - 25*C*a^5*b^4 + 9*B*a^4*b^5 + 20*C*a^3*b^6 - 6*B*a^2*b^7 - 4*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 (B + C \sec{\left (c + d x \right )}\right ) \sec ^{4}{\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*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)
[Out]
Integral((B + C*sec(c + d*x))*sec(c + d*x)**4/(a + b*sec(c + d*x))**3, x)
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Giac [B] time = 1.37067, size = 784, normalized size = 2.71 \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*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")
[Out]
((6*C*a^6 - 2*B*a^5*b - 15*C*a^4*b^2 + 5*B*a^3*b^3 + 12*C*a^2*b^4 - 6*B*a*b^5)*(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)) - (4*C*a^6*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 5*C*
a^5*b*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 5*B*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 - 6*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*C*
a^6*tan(1/2*d*x + 1/2*c) + 2*B*a^5*b*tan(1/2*d*x + 1/2*c) - 5*C*a^5*b*tan(1/2*d*x + 1/2*c) + 3*B*a^4*b^2*tan(1
/2*d*x + 1/2*c) + 7*C*a^4*b^2*tan(1/2*d*x + 1/2*c) - 5*B*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) - 6*B*a^2*b^4*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*tan(
1/2*d*x + 1/2*c)^2 - a - b)^2) - (3*C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (3*C*a - B*b)*log(abs(
tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b^3))/d