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

Optimal. Leaf size=359 \[ \frac{\left (16 a^2 A b+8 a^3 B+11 a b^2 B+4 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (8 a^2 A-9 a b B-4 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{4 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{b \left (15 a^2 B+20 a A b+4 b^2 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{b (7 a B+4 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{4 d \sqrt{\cos (c+d x)}}+\frac{b B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d \sqrt{\cos (c+d x)}} \]

[Out]

((16*a^2*A*b + 4*A*b^3 + 8*a^3*B + 11*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)
/(a + b)])/(4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (b*(20*a*A*b + 15*a^2*B + 4*b^2*B)*Sqrt[(b + a*
Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d
*x]]) + ((8*a^2*A - 4*A*b^2 - 9*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec
[c + d*x]])/(4*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (b*(4*A*b + 7*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x
])/(4*d*Sqrt[Cos[c + d*x]]) + (b*B*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 1.42883, antiderivative size = 359, 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, 4026, 4096, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{\left (16 a^2 A b+8 a^3 B+11 a b^2 B+4 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (8 a^2 A-9 a b B-4 A b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{4 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{b \left (15 a^2 B+20 a A b+4 b^2 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{b (7 a B+4 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{4 d \sqrt{\cos (c+d x)}}+\frac{b B \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{2 d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]

[Out]

((16*a^2*A*b + 4*A*b^3 + 8*a^3*B + 11*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)
/(a + b)])/(4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (b*(20*a*A*b + 15*a^2*B + 4*b^2*B)*Sqrt[(b + a*
Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d
*x]]) + ((8*a^2*A - 4*A*b^2 - 9*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec
[c + d*x]])/(4*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (b*(4*A*b + 7*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x
])/(4*d*Sqrt[Cos[c + d*x]]) + (b*B*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d*Sqrt[Cos[c + d*x]])

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 4026

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

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 \sqrt{\cos (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))}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt{\cos (c+d x)}}+\frac{1}{2} \left (\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 (4 a A-b B)+\left (4 a A b+2 a^2 B+b^2 B\right ) \sec (c+d x)+\frac{1}{2} b (4 A b+7 a B) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{b (4 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d \sqrt{\cos (c+d x)}}+\frac{b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt{\cos (c+d x)}}+\frac{1}{2} \left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a \left (8 a^2 A-4 A b^2-9 a b B\right )+\frac{1}{2} a \left (12 a A b+4 a^2 B+b^2 B\right ) \sec (c+d x)+\frac{1}{4} b \left (20 a A b+15 a^2 B+4 b^2 B\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{b (4 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d \sqrt{\cos (c+d x)}}+\frac{b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt{\cos (c+d x)}}+\frac{1}{2} \left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} a \left (8 a^2 A-4 A b^2-9 a b B\right )+\frac{1}{2} a \left (12 a A b+4 a^2 B+b^2 B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{8} \left (b \left (20 a A b+15 a^2 B+4 b^2 B\right ) \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 (4 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d \sqrt{\cos (c+d x)}}+\frac{b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt{\cos (c+d x)}}+\frac{1}{8} \left (\left (8 a^2 A-4 A b^2-9 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}{8} \left (\left (-16 a^2 A b-4 A b^3-8 a^3 B-11 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 \left (20 a A b+15 a^2 B+4 b^2 B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{8 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ &=\frac{b (4 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d \sqrt{\cos (c+d x)}}+\frac{b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt{\cos (c+d x)}}-\frac{\left (\left (-16 a^2 A b-4 A b^3-8 a^3 B-11 a b^2 B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{8 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (b \left (20 a A b+15 a^2 B+4 b^2 B\right ) \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}{8 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (8 a^2 A-4 A b^2-9 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}{8 \sqrt{b+a \cos (c+d x)}}\\ &=\frac{b \left (20 a A b+15 a^2 B+4 b^2 B\right ) \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 )}{4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{b (4 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d \sqrt{\cos (c+d x)}}+\frac{b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt{\cos (c+d x)}}-\frac{\left (\left (-16 a^2 A b-4 A b^3-8 a^3 B-11 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}{8 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (8 a^2 A-4 A b^2-9 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}{8 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=\frac{\left (16 a^2 A b+4 A b^3+8 a^3 B+11 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 )}{4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{b \left (20 a A b+15 a^2 B+4 b^2 B\right ) \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 )}{4 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (8 a^2 A-4 A b^2-9 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)}}{4 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{b (4 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{4 d \sqrt{\cos (c+d x)}}+\frac{b B (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{2 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 33.9063, size = 97208, normalized size = 270.77 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(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 = 0.605, size = 2216, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x)

[Out]

-1/4/d*(-1+cos(d*x+c))*(cos(d*x+c)+1)*(-9*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+8*
A*sin(d*x+c)*cos(d*x+c)^2*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^3-2*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(
1/2)-8*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(d*x+c)+1))^(1/2)-8*A*sin(d*x+c)*cos(d*x+c)^2*(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^3-4*A*sin(d*x+c)*cos(d*x+c)^2*(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^3-8*B*sin(d*x+c)*cos(d*x+c)^2*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+4*
B*sin(d*x+c)*cos(d*x+c)^2*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^3-8*B*sin(d*x+c)*cos(d*x+c)^2*(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+
11*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)-8*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b*
(1/(cos(d*x+c)+1))^(1/2)+4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+9*B*cos(d*x+c)^2*
((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+4*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^3*(1/(cos(d*x+c)+1))^(
1/2)-9*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+6*B*sin(d*x+c)*cos(d*x+c)^2*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*B*sin(d*x+c)*cos(d*x+c)^2*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^2-30*B*sin(d*x+c)*cos(d*x+c)^2*(1/(a+b)*(b+a*co
s(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^2*b+9*B*sin(d*x+c)*cos(d*x+c)^2*(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-9*B*sin(d*x+c)*cos(d*x+c)^2*(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^2-24*A*sin(d*x+c)*cos(d*x+c)^2*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+16*A*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+co
s(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^2-40*A*sin(d*x+c)*cos(d*x+c)^2*(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+8*A*sin(d*x+c)*cos(d*x+c)^2*(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+4*A*sin(d*x+c)*cos(d*x+c)^2*(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+8*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b*(1
/(cos(d*x+c)+1))^(1/2)-4*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+2*B*((a-b)/(a+b))^(
1/2)*b^3*(1/(cos(d*x+c)+1))^(1/2)+8*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^3-4*A*cos(d*
x+c)^2*((a-b)/(a+b))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*b^3-2*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*b^3*(1/(cos(d*x+c
)+1))^(1/2))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/(1/(cos(d*x+c)+1))^(1/2)
/cos(d*x+c)^(3/2)/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}} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\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((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sq
rt(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((a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))*cos(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 (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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