3.1336 \(\int \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=273 \[ -\frac{2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \sqrt{\cos (c+d x)} \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 (5 a B+A b) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 a d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{5 d} \]
[Out]
(-2*(a^2 - b^2)*(2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(15*
a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2*(2*A*b^2 - 5*a*b*B - 3*a^2*(3*A + 5*C))*Sqrt[Cos[c + d
*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(15*a^2*d*Sqrt[(b + a*Cos[c + d*x])/(a +
b)]) + (2*(A*b + 5*a*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*a*d) + (2*A*Cos[c + d*x]
^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d)
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Rubi [A] time = 0.938143, antiderivative size = 273, normalized size of antiderivative
= 1., number of steps used = 10, number of rules used = 10, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.222, Rules used = {4265, 4094, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ -\frac{2 \sqrt{\cos (c+d x)} \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}-\frac{2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 (5 a B+A b) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 a d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(5/2)*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(-2*(a^2 - b^2)*(2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(15*
a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2*(2*A*b^2 - 5*a*b*B - 3*a^2*(3*A + 5*C))*Sqrt[Cos[c + d
*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(15*a^2*d*Sqrt[(b + a*Cos[c + d*x])/(a +
b)]) + (2*(A*b + 5*a*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*a*d) + (2*A*Cos[c + d*x]
^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d)
Rule 4265
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]
Rule 4094
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Rule 4104
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Rule 4035
Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
Rule 3856
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2655
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 3858
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2663
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 2661
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rubi steps
\begin{align*} \int \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} (A b+5 a B)+\frac{1}{2} (3 a A+5 b B+5 a C) \sec (c+d x)+\frac{1}{2} b (2 A+5 C) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 (A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} \left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right )-\frac{1}{4} a (7 A b+5 a B+15 b C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 (A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}-\frac{\left (\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^2}-\frac{\left (\left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 (A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}-\frac{\left (\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{15 a^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{15 a^2 \sqrt{b+a \cos (c+d x)}}\\ &=\frac{2 (A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}-\frac{\left (\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{15 a^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{15 a^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=-\frac{2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) \sqrt{\cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{15 a^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{2 (A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}\\ \end{align*}
Mathematica [C] time = 17.7665, size = 404, normalized size = 1.48 \[ \frac{\sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \left (\frac{2 (5 a B+A b) \sin (c+d x)}{15 a}+\frac{1}{5} A \sin (2 (c+d x))\right )}{d}-\frac{2 \cos ^{\frac{3}{2}}(c+d x) \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \sqrt{a+b \sec (c+d x)} \left (i a (a+b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (9 a A+5 a (B+3 C)-2 A b) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )-\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} \left (3 a^2 (3 A+5 C)+5 a b B-2 A b^2\right ) (a \cos (c+d x)+b)-i (a+b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (3 a^2 (3 A+5 C)+5 a b B-2 A b^2\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{15 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+b)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(5/2)*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*((2*(A*b + 5*a*B)*Sin[c + d*x])/(15*a) + (A*Sin[2*(c + d*x)])/5))
/d - (2*Cos[c + d*x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*Sqrt[a + b*Sec[c + d*x]]*((-I)*(a + b)*(-2*
A*b^2 + 5*a*b*B + 3*a^2*(3*A + 5*C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]
^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(9*a*A - 2*A*b + 5*a*(B + 3*C))*Ellip
ticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x
)/2]^2)/(a + b)] - (-2*A*b^2 + 5*a*b*B + 3*a^2*(3*A + 5*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Ta
n[(c + d*x)/2]))/(15*a^2*d*(b + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]])
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Maple [B] time = 0.552, size = 1966, normalized size = 7.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x)
[Out]
-2/15/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^3*(3*A*sin(d*x+c
)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^3*(1/(cos(d*x+c)+1))^(3/2)+9*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(c
os(d*x+c)+1))^(3/2)+A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(3/2)+5*B*sin(d*x+c)*((a-b)/(a+b
))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(3/2)+5*B*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(3/2)+15*C
*((a-b)/(a+b))^(1/2)*a^2*b*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)+3*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2
*a^3*(1/(cos(d*x+c)+1))^(3/2)+9*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*(1/(cos(d*x+c)+1))^(3/2)+5*B*s
in(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^3*(1/(cos(d*x+c)+1))^(3/2)+5*B*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos
(d*x+c)*a^3*(1/(cos(d*x+c)+1))^(3/2)-2*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x
+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^3+9*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2
)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3-9*A*(1/(a+b)*(b+a*cos(d
*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^
3-5*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))
/(cos(d*x+c)+1))^(1/2)*a^3+15*C*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))
*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3-15*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Elli
pticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3-2*A*sin(d*x+c)*((a-b)/(a+b))^(1
/2)*b^3*(1/(cos(d*x+c)+1))^(3/2)+5*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(
1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b-5*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2
)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b+5*B*(1/(a+b)*(b+a*cos(d
*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*
b^2-15*C*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)*a^2*b+15*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))
*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-7*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/
sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b-2*A*EllipticF((-1+cos(d
*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*
b^2+9*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+
c),(-(a+b)/(a-b))^(1/2))*a^2*b+2*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*(
(a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+4*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2*
b*(1/(cos(d*x+c)+1))^(3/2)+4*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b*(1/(cos(d*x+c)+1))^(3/2)-A*sin(
d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^2*(1/(cos(d*x+c)+1))^(3/2)+10*B*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d
*x+c)*a^2*b*(1/(cos(d*x+c)+1))^(3/2)+15*C*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/
2))/a^2/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/(1/(cos(d*x+c)+1))^(3/2)/sin(d*x+c)^6
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*cos(d*x + c)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)^2*sec(d*x + c)^2 + B*cos(d*x + c)^2*sec(d*x + c) + A*cos(d*x + c)^2)*sqrt(b*sec(d*x +
c) + a)*sqrt(cos(d*x + c)), x)
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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(cos(d*x+c)**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+b*sec(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*cos(d*x + c)^(5/2), x)