\(\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1336]
Optimal result
Integrand size = 45, antiderivative size = 273 \[
\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 (A b+5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}
\]
[Out]
-2/15*(a^2-b^2)*(2*A*b-5*B*a)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(
1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/a^2/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/5*A*cos(d
*x+c)^(3/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/d+2/15*(A*b+5*B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^
(1/2)/a/d-2/15*(2*A*b^2-5*B*a*b-3*a^2*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin
(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^2/d/((b+a*cos(d*x+c))/(a+b)
)^(1/2)
Rubi [A] (verified)
Time = 1.09 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of
steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4350, 4179, 4189, 4120,
3941, 2734, 2732, 3943, 2742, 2740} \[
\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \sqrt {\cos (c+d x)} \left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^2 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 a d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 d}
\]
[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 2732
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2734
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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2740
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2742
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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 3941
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 3943
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[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 4120
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 4179
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 4189
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 4350
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]
Rubi steps \begin{align*}
\text {integral}& = \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}} \operatorname {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 \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] (warning: unable to verify)
Result contains complex when optimal does not.
Time = 36.14 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.48
\[
\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \left (\frac {2 (A b+5 a B) \sin (c+d x)}{15 a}+\frac {1}{5} A \sin (2 (c+d x))\right )}{d}-\frac {2 \cos ^{\frac {3}{2}}(c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \sqrt {a+b \sec (c+d x)} \left (-i (a+b) \left (-2 A b^2+5 a b B+3 a^2 (3 A+5 C)\right ) E\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+i a (a+b) (9 a A-2 A b+5 a (B+3 C)) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-\left (-2 A b^2+5 a b B+3 a^2 (3 A+5 C)\right ) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 a^2 d (b+a \cos (c+d x)) \sqrt {\sec (c+d x)}}
\]
[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]])
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(3192\) vs. \(2(303)=606\).
Time = 6.38 (sec) , antiderivative size = 3193, normalized size of antiderivative =
11.70
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| method | result | size |
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\(\text {Expression too large to display}\) |
\(3193\) |
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[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,method=_RETURNVERBOSE)
[Out]
2/15/d*(3*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*(1/(1+cos(d*x+c)))^(1/2)+15*C*(1/(a+b)*(b+a*cos(d*
x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b*
cos(d*x+c)-15*C*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(
d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b*cos(d*x+c)+5*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^3*cos(d*x+c)
^2*sin(d*x+c)+3*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*(1/(1+cos(d*x+c)))^(1/2)+9*A*cos(d*x+c)*((a-
b)/(a+b))^(1/2)*sin(d*x+c)*a^3*(1/(1+cos(d*x+c)))^(1/2)+15*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+
c)))^(1/2)*sin(d*x+c)+9*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+A*((a-b)/(a+b))^(1/2)*
sin(d*x+c)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)+5*B*sin(d*x+c)*cos(d*x+c)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/
2)*a^3+2*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c
)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2-7*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ellip
ticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b-2*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(
d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^
2+4*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+9*A*cos(d*x+c)*(1/(a+b)*(b+a*co
s(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^
2*b+5*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+5*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a*b^2
*(1/(1+cos(d*x+c)))^(1/2)+15*C*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-5*B*(1/(a+b)*(b+a
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))
*a^2*b*cos(d*x+c)+5*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c
)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2*cos(d*x+c)+5*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ellip
ticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b*cos(d*x+c)-2*A*((a-b)/(a+b))^(1/2
)*sin(d*x+c)*b^3*(1/(1+cos(d*x+c)))^(1/2)-2*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)
/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2-5*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c))
)^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*cos(d*x+c)-15*C*(1/(a+
b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))
^(1/2))*a^3*cos(d*x+c)-9*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^
(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3-2*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)
))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b^3+9*A*cos(d*x+c)*(1/(a+
b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))
^(1/2))*a^3+9*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(
d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b+2*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b)
)^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2-7*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2
)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b-5*B*(1/(a+b)*(b+a*cos(d*x+
c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b+5*
B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b
)/(a-b))^(1/2))*a*b^2+5*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d
*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b+15*C*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE((
(a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b-15*C*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d
*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b+15*C*(1/(a+b)*
(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1
/2))*a^3*cos(d*x+c)-A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)+10*B*cos(d*x+c)
*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x+c)
*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-5*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1
/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3-15*C*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ell
ipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3+15*C*(1/(a+b)*(b+a*cos(d*x+c))/(1
+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3-9*A*(1/(a+
b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))
^(1/2))*a^3-2*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(
d*x+c)),(-(a+b)/(a-b))^(1/2))*b^3+9*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^
(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/(1/(1+cos(d*x
+c)))^(1/2)/(b+a*cos(d*x+c))/((a-b)/(a+b))^(1/2)/a^2/(1+cos(d*x+c))
Fricas [C] (verification not implemented)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.94
\[
\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=\frac {6 \, {\left (3 \, A a^{3} \cos \left (d x + c\right ) + 5 \, B a^{3} + A a^{2} b\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-15 i \, B a^{3} - 3 i \, {\left (A + 5 \, C\right )} a^{2} b + 10 i \, B a b^{2} - 4 i \, A b^{3}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) + \sqrt {2} {\left (15 i \, B a^{3} + 3 i \, {\left (A + 5 \, C\right )} a^{2} b - 10 i \, B a b^{2} + 4 i \, A b^{3}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) - 3 \, \sqrt {2} {\left (-3 i \, {\left (3 \, A + 5 \, C\right )} a^{3} - 5 i \, B a^{2} b + 2 i \, A a b^{2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, {\left (3 \, A + 5 \, C\right )} a^{3} + 5 i \, B a^{2} b - 2 i \, A a b^{2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right )}{45 \, a^{3} d}
\]
[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]
1/45*(6*(3*A*a^3*cos(d*x + c) + 5*B*a^3 + A*a^2*b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x + c))*
sin(d*x + c) + sqrt(2)*(-15*I*B*a^3 - 3*I*(A + 5*C)*a^2*b + 10*I*B*a*b^2 - 4*I*A*b^3)*sqrt(a)*weierstrassPInve
rse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)
+ sqrt(2)*(15*I*B*a^3 + 3*I*(A + 5*C)*a^2*b - 10*I*B*a*b^2 + 4*I*A*b^3)*sqrt(a)*weierstrassPInverse(-4/3*(3*a
^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 3*sqrt(2)*
(-3*I*(3*A + 5*C)*a^3 - 5*I*B*a^2*b + 2*I*A*a*b^2)*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a
^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x
+ c) + 3*I*a*sin(d*x + c) + 2*b)/a)) - 3*sqrt(2)*(3*I*(3*A + 5*C)*a^3 + 5*I*B*a^2*b - 2*I*A*a*b^2)*sqrt(a)*wei
erstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^
2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a)))/(a^3*d)
Sympy [F(-1)]
Timed out. \[
\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out}
\]
[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
Maxima [F]
\[
\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x }
\]
[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)
Giac [F]
\[
\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x }
\]
[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)
Mupad [F(-1)]
Timed out. \[
\int \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x
\]
[In]
int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)