3.401 \(\int \frac{\sec ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=261 \[ -\frac{(85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{80 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(163 B-283 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(475 B-787 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{240 a^3 d}-\frac{(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac{(13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
[Out]
((163*B - 283*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (
(B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((13*B - 21*C)*Sec[c + d*x]^3*Tan[c +
d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((985*B - 1729*C)*Tan[c + d*x])/(120*a^2*d*Sqrt[a + a*Sec[c + d*x]
]) - ((85*B - 157*C)*Sec[c + d*x]^2*Tan[c + d*x])/(80*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((475*B - 787*C)*Sqrt[
a + a*Sec[c + d*x]]*Tan[c + d*x])/(240*a^3*d)
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Rubi [A] time = 0.928138, antiderivative size = 261, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{4072, 4019, 4021, 4010, 4001, 3795, 203} \[ -\frac{(85 B-157 C) \tan (c+d x) \sec ^2(c+d x)}{80 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(163 B-283 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(475 B-787 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{240 a^3 d}-\frac{(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(B-C) \tan (c+d x) \sec ^4(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac{(13 B-21 C) \tan (c+d x) \sec ^3(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]
[Out]
((163*B - 283*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (
(B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((13*B - 21*C)*Sec[c + d*x]^3*Tan[c +
d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((985*B - 1729*C)*Tan[c + d*x])/(120*a^2*d*Sqrt[a + a*Sec[c + d*x]
]) - ((85*B - 157*C)*Sec[c + d*x]^2*Tan[c + d*x])/(80*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((475*B - 787*C)*Sqrt[
a + a*Sec[c + d*x]]*Tan[c + d*x])/(240*a^3*d)
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 4019
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0]
Rule 4021
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(f*(m + n
)), x] + Dist[d/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m +
n) + a*B*m)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b
^2, 0] && GtQ[n, 1]
Rule 4010
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && !LtQ[m, -1]
Rule 4001
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && !LtQ[m, -2^(-1)]
Rule 3795
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 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 \frac{\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\int \frac{\sec ^5(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\sec ^4(c+d x) \left (4 a (B-C)-\frac{1}{2} a (5 B-13 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec ^3(c+d x) \left (\frac{3}{2} a^2 (13 B-21 C)-\frac{1}{4} a^2 (85 B-157 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(85 B-157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\sec ^2(c+d x) \left (-\frac{1}{2} a^3 (85 B-157 C)+\frac{1}{8} a^3 (475 B-787 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{20 a^5}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(85 B-157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(475 B-787 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}+\frac{\int \frac{\sec (c+d x) \left (\frac{1}{16} a^4 (475 B-787 C)-\frac{1}{8} a^4 (985 B-1729 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{30 a^6}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(85 B-157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(475 B-787 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}+\frac{(163 B-283 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(85 B-157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(475 B-787 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}-\frac{(163 B-283 C) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{(163 B-283 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(B-C) \sec ^4(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(13 B-21 C) \sec ^3(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(985 B-1729 C) \tan (c+d x)}{120 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(85 B-157 C) \sec ^2(c+d x) \tan (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(475 B-787 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{240 a^3 d}\\ \end{align*}
Mathematica [A] time = 2.60335, size = 177, normalized size = 0.68 \[ \frac{\tan (c+d x) \left (\sqrt{1-\sec (c+d x)} \left (160 (B-C) \sec ^3(c+d x)-32 (25 B-49 C) \sec ^2(c+d x)-5 (503 B-911 C) \sec (c+d x)-1495 B+96 C \sec ^4(c+d x)+2671 C\right )+30 \sqrt{2} (163 B-283 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{240 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sec[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]
[Out]
((30*Sqrt[2]*(163*B - 283*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2 + Sqrt[
1 - Sec[c + d*x]]*(-1495*B + 2671*C - 5*(503*B - 911*C)*Sec[c + d*x] - 32*(25*B - 49*C)*Sec[c + d*x]^2 + 160*(
B - C)*Sec[c + d*x]^3 + 96*C*Sec[c + d*x]^4))*Tan[c + d*x])/(240*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]
))^(5/2))
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Maple [B] time = 0.325, size = 985, normalized size = 3.8 \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)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)
[Out]
1/1920/d/a^3*(-1+cos(d*x+c))^2*(2445*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^4-4245*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2
)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^4+9780*B
*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x
+c)-cos(d*x+c)+1)/sin(d*x+c))-16980*C*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*ln(((-2*cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+14670*B*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)-25470*
C*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(5/2)*sin(d*x+c)+9780*B*cos(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c
)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)-16980*C*cos(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+2445*B*ln(
((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2
)*sin(d*x+c)-4245*C*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+11960*B*cos(d*x+c)^5-21368*C*cos(d*x+c)^5+8160*B*cos(d*x+c)^4-15072*C*cos(
d*x+c)^4-13720*B*cos(d*x+c)^3+23896*C*cos(d*x+c)^3-7680*B*cos(d*x+c)^2+13824*C*cos(d*x+c)^2+1280*B*cos(d*x+c)-
2048*C*cos(d*x+c)+768*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(d*x+c)^2
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [A] time = 0.686998, size = 1597, normalized size = 6.12 \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)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
[1/960*(15*sqrt(2)*((163*B - 283*C)*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos(d*x + c)^4 + 3*(163*B - 283*C)*cos(
d*x + c)^3 + (163*B - 283*C)*cos(d*x + c)^2)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x +
c) + 1)) - 4*((1495*B - 2671*C)*cos(d*x + c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 32*(25*B - 49*C)*cos(d*x +
c)^2 - 160*(B - C)*cos(d*x + c) - 96*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x
+ c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2), -1/480*(15*sqrt(2)*((163*B -
283*C)*cos(d*x + c)^5 + 3*(163*B - 283*C)*cos(d*x + c)^4 + 3*(163*B - 283*C)*cos(d*x + c)^3 + (163*B - 283*C)
*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x
+ c))) + 2*((1495*B - 2671*C)*cos(d*x + c)^4 + 5*(503*B - 911*C)*cos(d*x + c)^3 + 32*(25*B - 49*C)*cos(d*x + c
)^2 - 160*(B - C)*cos(d*x + c) - 96*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x +
c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [A] time = 10.0303, size = 593, normalized size = 2.27 \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)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
-1/480*((((15*(2*(sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
))*tan(1/2*d*x + 1/2*c)^2/a^2 + (21*sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 29*sqrt(2)*C*a^2*sgn(tan(1
/2*d*x + 1/2*c)^2 - 1))/a^2)*tan(1/2*d*x + 1/2*c)^2 - (3685*sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 67
33*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^2)*tan(1/2*d*x + 1/2*c)^2 + 5*(1133*sqrt(2)*B*a^2*sgn(tan(
1/2*d*x + 1/2*c)^2 - 1) - 1973*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^2)*tan(1/2*d*x + 1/2*c)^2 - 15
*(155*sqrt(2)*B*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 291*sqrt(2)*C*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^2)*
tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)) - 15*(163*sqrt(2)*
B - 283*sqrt(2)*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^
2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d