3.1351 \(\int \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=419 \[ \frac{\left (4 a^2 b (4 A+15 C)+10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\sqrt{\cos (c+d x)} \left (6 a^2 (3 A+5 C)+70 a b B+b^2 (46 A-15 C)\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}-\frac{b \sin (c+d x) (10 a B+16 A b-15 b C) \sqrt{a+b \sec (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (a B+A b) \sin (c+d x) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}{3 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}{5 d}+\frac{b^2 (5 a C+2 b B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]
[Out]
((10*a^3*B + 20*a*b^2*B - b^3*(16*A - 15*C) + 4*a^2*b*(4*A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*a)/(a + b)])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (b^2*(2*b*B + 5*a*C)*Sqrt
[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec
[c + d*x]]) + ((70*a*b*B + b^2*(46*A - 15*C) + 6*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*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (b*(16*A*b + 10*a*B - 15*b*
C)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(A*b + a*B)*Sqrt[Cos[c + d*x]]*(a + b
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*A*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5
*d)
________________________________________________________________________________________
Rubi [A] time = 1.828, antiderivative size = 419, normalized size of antiderivative = 1.,
number of steps used = 15, number of rules used = 14, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.311, Rules used
= {4265, 4094, 4096, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac{\left (4 a^2 b (4 A+15 C)+10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\sqrt{\cos (c+d x)} \left (6 a^2 (3 A+5 C)+70 a b B+b^2 (46 A-15 C)\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}-\frac{b \sin (c+d x) (10 a B+16 A b-15 b C) \sqrt{a+b \sec (c+d x)}}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (a B+A b) \sin (c+d x) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}{3 d}+\frac{2 A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}{5 d}+\frac{b^2 (5 a C+2 b B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
((10*a^3*B + 20*a*b^2*B - b^3*(16*A - 15*C) + 4*a^2*b*(4*A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*a)/(a + b)])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (b^2*(2*b*B + 5*a*C)*Sqrt
[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec
[c + d*x]]) + ((70*a*b*B + b^2*(46*A - 15*C) + 6*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*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (b*(16*A*b + 10*a*B - 15*b*
C)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*(A*b + a*B)*Sqrt[Cos[c + d*x]]*(a + b
*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + (2*A*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)*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 4096
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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(m + n + 1), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^
n*Simp[a*A*(m + n + 1) + a*C*n + ((A*b + a*B)*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) + a*C
*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] &&
!LeQ[n, -1]
Rule 4108
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Rule 3859
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2807
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
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) (a+b \sec (c+d x))^{5/2} \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{(a+b \sec (c+d x))^{5/2} \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) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \sec (c+d x))^{3/2} \left (\frac{5}{2} (A b+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)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 (A b+a B) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{15} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)} \left (\frac{3}{4} \left (5 A b^2+10 a b B+a^2 (3 A+5 C)\right )+\frac{1}{4} \left (8 a A b+5 a^2 B+15 b^2 B+30 a b C\right ) \sec (c+d x)-\frac{1}{4} b (16 A b+10 a B-15 b C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (A b+a B) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{15} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{8} a \left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right )+\frac{1}{4} \left (15 A b^3+5 a^3 B+45 a b^2 B+a^2 b (17 A+45 C)\right ) \sec (c+d x)+\frac{15}{8} b^2 (2 b B+5 a C) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (A b+a B) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{15} \left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{8} a \left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right )+\frac{1}{4} \left (15 A b^3+5 a^3 B+45 a b^2 B+a^2 b (17 A+45 C)\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{2} \left (b^2 (2 b B+5 a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (A b+a B) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}+\frac{1}{30} \left (\left (70 a b B+b^2 (46 A-15 C)+6 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-\frac{1}{30} \left (\left (-10 a^3 B-20 a b^2 B+b^3 (16 A-15 C)-4 a^2 b (4 A+15 C)\right ) \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+\frac{\left (b^2 (2 b B+5 a C) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (A b+a B) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac{\left (\left (-10 a^3 B-20 a b^2 B+b^3 (16 A-15 C)-4 a^2 b (4 A+15 C)\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{30 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (b^2 (2 b B+5 a C) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (70 a b B+b^2 (46 A-15 C)+6 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}{30 \sqrt{b+a \cos (c+d x)}}\\ &=\frac{b^2 (2 b B+5 a C) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{b (16 A b+10 a B-15 b C) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (A b+a B) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}-\frac{\left (\left (-10 a^3 B-20 a b^2 B+b^3 (16 A-15 C)-4 a^2 b (4 A+15 C)\right ) \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}{30 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (70 a b B+b^2 (46 A-15 C)+6 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}{30 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=\frac{\left (10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)+4 a^2 b (4 A+15 C)\right ) \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 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{b^2 (2 b B+5 a C) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (70 a b B+b^2 (46 A-15 C)+6 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 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}-\frac{b (16 A b+10 a B-15 b C) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d \sqrt{\cos (c+d x)}}+\frac{2 (A b+a B) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{2 A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d}\\ \end{align*}
Mathematica [C] time = 34.8934, size = 86542, normalized size = 206.54 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
Result too large to show
________________________________________________________________________________________
Maple [C] time = 0.706, size = 2893, normalized size = 6.9 \begin{align*} \text{output 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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
-1/15/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^3*(15*C*cos(d*x+c)*((a-b)/(a+b))^
(1/2)*a*b^2*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)+6*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)+18*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)+10*B*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)+28*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*
a^2*b*(1/(cos(d*x+c)+1))^(3/2)+68*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a*b^2*(1/(cos(d*x+c)+1))^(3/2)
+80*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^2*b*(1/(cos(d*x+c)+1))^(3/2)+70*B*((a-b)/(a+b))^(1/2)*cos(
d*x+c)*sin(d*x+c)*a*b^2*(1/(cos(d*x+c)+1))^(3/2)+30*C*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b*(1/(cos(d*x+c)+1))^
(3/2)*sin(d*x+c)+15*C*((a-b)/(a+b))^(1/2)*b^3*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)+28*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)+18*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b*(1/(c
os(d*x+c)+1))^(3/2)+22*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)+18*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(c
os(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3-30*A*co
s(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x
+c),(-(a+b)/(a-b))^(1/2))*b^3-18*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+co
s(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3+46*A*cos(d*x+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))*b^3-10*B*c
os(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*
x+c),(-(a+b)/(a-b))^(1/2))*a^3+30*B*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+c
os(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^3-60*B*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/
(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1
/2))*b^3+30*C*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b
))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3-30*C*cos(d*x+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^3-15*C*cos(d*x+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))*b^3+46*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*sin(d*x+c)*b^3*(1/(cos(d*x+c)+1))^(3/2)+6*A*((a-b)/(a+b))^(1/2)*
cos(d*x+c)^4*sin(d*x+c)*a^3*(1/(cos(d*x+c)+1))^(3/2)+70*B*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))
^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-90*B*cos(d*x+c)*(1
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)
/(a-b))^(1/2))*a*b^2-70*B*cos(d*x+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+70*B*cos(d*x+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*b^2-90*C*cos(d*
x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),
(-(a+b)/(a-b))^(1/2))*a^2*b+60*C*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(
d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+30*C*cos(d*x+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+15*C
*cos(d*x+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*b^2-150*C*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi
((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a*b^2-34*A*cos(d*x+c)*(1/(a
+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a
-b))^(1/2))*a^2*b+46*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((
a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2+18*A*cos(d*x+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-46*A*cos(d*x+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*b^2+10*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^3*(1/(cos(d*x+c)+1))^(3/2)+30*C*((
a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^3*(1/(cos(d*x+c)+1))^(3/2)*sin(d*x+c))/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/s
in(d*x+c)^6/(1/(cos(d*x+c)+1))^(3/2)/cos(d*x+c)^(1/2)
________________________________________________________________________________________
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(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
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))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
________________________________________________________________________________________
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 )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2), x)