3.610 \(\int \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=342 \[ \frac{2 \left (8 a^2 A b+5 a^3 B+10 a b^2 B-8 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2 A+35 a b B+23 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a (5 a B+8 A b) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}+\frac{2 b^3 B \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]

[Out]

(2*(8*a^2*A*b - 8*A*b^3 + 5*a^3*B + 10*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a
)/(a + b)])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*b^3*B*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*E
llipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^2*A + 23*A
*b^2 + 35*a*b*B)*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)]) + (2*a*(8*A*b + 5*a*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x
])/(15*d) + (2*a*A*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 1.39308, antiderivative size = 342, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 14, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2955, 4025, 4094, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \left (8 a^2 A b+5 a^3 B+10 a b^2 B-8 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2 A+35 a b B+23 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a (5 a B+8 A b) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 a A \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}+\frac{2 b^3 B \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]

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]),x]

[Out]

(2*(8*a^2*A*b - 8*A*b^3 + 5*a^3*B + 10*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a
)/(a + b)])/(15*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*b^3*B*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*E
llipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^2*A + 23*A
*b^2 + 35*a*b*B)*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)]) + (2*a*(8*A*b + 5*a*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x
])/(15*d) + (2*a*A*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

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 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} (A+B \sec (c+d x)) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)} \left (-\frac{1}{2} a (8 A b+5 a B)-\frac{1}{2} \left (3 a^2 A+5 A b^2+10 a b B\right ) \sec (c+d x)-\frac{5}{2} b^2 B \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a (8 A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \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}{4} a \left (9 a^2 A+23 A b^2+35 a b B\right )-\frac{1}{4} \left (17 a^2 A b+15 A b^3+5 a^3 B+45 a b^2 B\right ) \sec (c+d x)-\frac{15}{4} b^3 B \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 a (8 A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \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}{4} a \left (9 a^2 A+23 A b^2+35 a b B\right )+\frac{1}{4} \left (-17 a^2 A b-15 A b^3-5 a^3 B-45 a b^2 B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx+\left (b^3 B \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{2 a (8 A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{1}{15} \left (\left (-9 a^2 A-23 A b^2-35 a b B\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}{15} \left (\left (-8 a^2 A b+8 A b^3-5 a^3 B-10 a b^2 B\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^3 B \sqrt{b+a \cos (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{\sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ &=\frac{2 a (8 A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{\left (\left (-8 a^2 A b+8 A b^3-5 a^3 B-10 a b^2 B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{15 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (b^3 B \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}{\sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (-9 a^2 A-23 A b^2-35 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{15 \sqrt{b+a \cos (c+d x)}}\\ &=\frac{2 b^3 B \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{2 a (8 A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{\left (\left (-8 a^2 A b+8 A b^3-5 a^3 B-10 a b^2 B\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}{15 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (-9 a^2 A-23 A b^2-35 a b B\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 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=\frac{2 \left (8 a^2 A b-8 A b^3+5 a^3 B+10 a b^2 B\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{2 b^3 B \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{2 \left (9 a^2 A+23 A b^2+35 a b B\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{2 a (8 A b+5 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a A \cos ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 34.447, size = 49609, normalized size = 145.06 \[ \text{Result too large to show} \]

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]),x]

[Out]

Result too large to show

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Maple [C]  time = 0.43, size = 2052, normalized size = 6. \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))^(5/2)*(A+B*sec(d*x+c)),x)

[Out]

2/15/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)+1)*(-23*A*sin(d*x+c)*E
llipticF((-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*sin(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+3*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(d
*x+c)+1))^(1/2)+6*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(d*x+c)+1))^(1/2)-9*A*sin(d*x+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+9*A*sin(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^3-23*A*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*El
lipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^3+23*A*sin(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+35*B*sin(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-35*B*sin(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^2-35*B*sin(d*x+c)*Ellipt
icF((-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+17*A*sin(d*x+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-5*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c
)+1))^(1/2)-23*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)-35*B*cos(d*x+c)*((a-b)/(a+b))^(
1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+35*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+14*A*co
s(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+34*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^2*(1/(
cos(d*x+c)+1))^(1/2)+40*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+5*B*cos(d*x+c)^3*(1/
(cos(d*x+c)+1))^(1/2)*((a-b)/(a+b))^(1/2)*a^3+5*B*sin(d*x+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^3-5*B*cos(d*x+c)*((a-b)/(a+b))
^(1/2)*a^3*(1/(cos(d*x+c)+1))^(1/2)-9*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(d*x+c)+1))^(1/2)+23*A*cos(d
*x+c)*((a-b)/(a+b))^(1/2)*b^3*(1/(cos(d*x+c)+1))^(1/2)-9*A*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)-
11*A*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)-5*B*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)
-35*B*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+15*A*sin(d*x+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)*b^3-15*B*sin(d*x+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)*b^3+30*B*sin(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^3-23*A*((a-b)/(a+b))^(1/2)*b^3*(1/(cos(d
*x+c)+1))^(1/2)+45*B*sin(d*x+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*b^2)/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/(1/(cos(d*x+c)+1)
)^(1/2)/sin(d*x+c)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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)),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2), x)

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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)),x, algorithm="fricas")

[Out]

Timed out

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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))**(5/2)*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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)),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2), x)