3.622 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=326 \[ \frac{2 \left (a^2 A-6 a b B+8 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A+3 a b B-4 A b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )}+\frac{2 b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{a d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (5 a^2 A b-3 a^3 B+6 a b^2 B-8 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(2*(a^2*A + 8*A*b^2 - 6*a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*a^
3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2*(5*a^2*A*b - 8*A*b^3 - 3*a^3*B + 6*a*b^2*B)*Sqrt[Cos[c +
d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^3*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c
+ d*x])/(a + b)]) + (2*b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]
]) + (2*(a^2*A - 4*A*b^2 + 3*a*b*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a^2*(a^2 - b^
2)*d)

________________________________________________________________________________________

Rubi [A]  time = 1.03192, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2955, 4030, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \left (a^2 A+3 a b B-4 A b^2\right ) \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{3 a^2 d \left (a^2-b^2\right )}+\frac{2 b (A b-a B) \sin (c+d x) \sqrt{\cos (c+d x)}}{a d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A-6 a b B+8 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (5 a^2 A b-3 a^3 B+6 a b^2 B-8 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*(a^2*A + 8*A*b^2 - 6*a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*a^
3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2*(5*a^2*A*b - 8*A*b^3 - 3*a^3*B + 6*a*b^2*B)*Sqrt[Cos[c +
d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^3*(a^2 - b^2)*d*Sqrt[(b + a*Cos[c
+ d*x])/(a + b)]) + (2*b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]
]) + (2*(a^2*A - 4*A*b^2 + 3*a*b*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a^2*(a^2 - b^
2)*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 4030

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

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 \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+B \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx\\ &=\frac{2 b (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} \left (-a^2 A+4 A b^2-3 a b B\right )+\frac{1}{2} a (A b-a B) \sec (c+d x)-b (A b-a B) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A-4 A b^2+3 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{4} \left (-5 a^2 A b+8 A b^3+3 a^3 B-6 a b^2 B\right )+\frac{1}{4} a \left (a^2 A+2 A b^2-3 a b B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A-4 A b^2+3 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (\left (a^2 A+8 A b^2-6 a b 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}{3 a^3}+\frac{\left (\left (-5 a^2 A b+8 A b^3+3 a^3 B-6 a b^2 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}{3 a^3 \left (a^2-b^2\right )}\\ &=\frac{2 b (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A-4 A b^2+3 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (\left (a^2 A+8 A b^2-6 a b B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 a^3 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (-5 a^2 A b+8 A b^3+3 a^3 B-6 a b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 a^3 \left (a^2-b^2\right ) \sqrt{b+a \cos (c+d x)}}\\ &=\frac{2 b (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A-4 A b^2+3 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac{\left (\left (a^2 A+8 A b^2-6 a b 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}{3 a^3 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (-5 a^2 A b+8 A b^3+3 a^3 B-6 a b^2 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}{3 a^3 \left (a^2-b^2\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=\frac{2 \left (a^2 A+8 A b^2-6 a b 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 )}{3 a^3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (5 a^2 A b-8 A b^3-3 a^3 B+6 a b^2 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)}}{3 a^3 \left (a^2-b^2\right ) d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{2 b (A b-a B) \sqrt{\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (a^2 A-4 A b^2+3 a b B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 17.014, size = 417, normalized size = 1.28 \[ \frac{2 (a \cos (c+d x)+b) \left (a \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (b \left (a^2 A+3 a b B-4 A b^2\right )+a A \left (a^2-b^2\right ) \cos (c+d x)\right )+\left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-i a \left (a^2-a b-2 b^2\right ) (a (A+3 B)-4 A b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )+\left (-5 a^2 A b+3 a^3 B-6 a b^2 B+8 A b^3\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)+i (a+b) \left (-5 a^2 A b+3 a^3 B-6 a b^2 B+8 A b^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )\right )}{3 a^3 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*(b + a*Cos[c + d*x])*(a*(b*(a^2*A - 4*A*b^2 + 3*a*b*B) + a*A*(a^2 - b^2)*Cos[c + d*x])*Sec[c + d*x]^(3/2)*S
in[c + d*x] + (Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(I*(a + b)*(-5*a^2*A*b + 8*A*b^3 + 3*a^3*B - 6*a*b^2*B)*
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^2 - a*b - 2*b^2)*(-4*A*b + a*(A + 3*B))*EllipticF[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)] + (-5*a^2*A*b + 8
*A*b^3 + 3*a^3*B - 6*a*b^2*B)*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2])))/(3*a^3*(a^2
- b^2)*d*Cos[c + d*x]^(3/2)*Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.342, size = 1460, normalized size = 4.5 \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)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x)

[Out]

2/3/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)^2*(8*A*sin(d*x+c)*El
lipticF((-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^2-5*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+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+8*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))*b^3-6*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^2-6*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^2*b+6*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+4*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+6*B*cos(d*x+c)*((a-b)
/(a+b))^(1/2)*a*b^2*(1/(cos(d*x+c)+1))^(1/2)+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)+3*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*
b*(1/(cos(d*x+c)+1))^(1/2)-3*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^3-3*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(
d*x+c)+1))^(1/2)-A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(d*x+c)+1))^(1/2)-8*A*cos(d*x+c)*((a-b)/(a+b))^(1
/2)*b^3*(1/(cos(d*x+c)+1))^(1/2)-A*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+4*A*((a-b)/(a+b))^(1/2)*
a*b^2*(1/(cos(d*x+c)+1))^(1/2)-3*B*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)-6*B*((a-b)/(a+b))^(1/2)*
a*b^2*(1/(cos(d*x+c)+1))^(1/2)-4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b*(1/(cos(d*x+c)+1))^(1/2)+8*A*((a-b)/
(a+b))^(1/2)*b^3*(1/(cos(d*x+c)+1))^(1/2)+A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^3+3*B*
cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*(1/(cos(d*x+c)+1))^(1/2)+3*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^3)*((a-b)/(a+b
))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)/a^3/(b+a*cos(d*x+c))/(a-b)/sin(d*x+c)^3

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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)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A \cos \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c))/(b^2*sec(d
*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), 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)**(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(b*sec(d*x + c) + a)^(3/2), x)