3.7.22 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [622]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.65, antiderivative size = 326, normalized size of antiderivative = 1.00, 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 = {3034, 4115, 4189, 4120, 3941, 2734, 2732, 3943, 2742, 2740} \begin {gather*} \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 (-3 a^3 B+5 a^2 A 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}}} \end {gather*}

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 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 3034

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 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 4115

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 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 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]

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] Result contains complex when optimal does not.
time = 11.41, size = 417, normalized size = 1.28 \begin {gather*} \frac {2 (b+a \cos (c+d x)) \left (a \left (b \left (a^2 A-4 A b^2+3 a b B\right )+a A \left (a^2-b^2\right ) \cos (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+\left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (i (a+b) \left (-5 a^2 A b+8 A b^3+3 a^3 B-6 a b^2 B\right ) E\left (i \sinh ^{-1}\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 \left (a^2-a b-2 b^2\right ) (-4 A b+a (A+3 B)) F\left (i \sinh ^{-1}\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 (-5 a^2 A b+8 A b^3+3 a^3 B-6 a b^2 B\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 )\right )}{3 a^3 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(1459\) vs. \(2(358)=716\).
time = 35.34, size = 1460, normalized size = 4.48

method	result	size
default	\(\text {Expression too large to display}\)	\(1460\)

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,method=_RETURNVERBOSE)

[Out]

2/3/d*cos(d*x+c)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(1+cos(d*x+c))^2*(8*A*((a-b)/(a+b))
^(1/2)*b^3*(1/(1+cos(d*x+c)))^(1/2)+A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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))*a^3+8*A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(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+A*((a-b)
/(a+b))^(1/2)*cos(d*x+c)^3*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-3*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*(1/(1+cos(d*x
+c)))^(1/2)-3*B*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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-A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*(1/(1+cos(d*x+c)))^(1/2)-8*
A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*b^3*(1/(1+cos(d*x+c)))^(1/2)-A*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(
1/2)+4*A*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)-3*B*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(
1/2)-6*B*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(1/2)+3*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos
(d*x+c)))^(1/2)-4*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-4*A*((a-b)/(a+b))^(1/2)*co
s(d*x+c)^2*a*b^2*(1/(1+cos(d*x+c)))^(1/2)+3*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+
4*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+6*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b^2*(1/
(1+cos(d*x+c)))^(1/2)+6*A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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+8*A*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b
))^(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^2-5*A*sin(d*x+c)*(
(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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-6*B*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))
^(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+A*cos(d*x+c)^3*((a
-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(1/2)+3*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+
cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^3)*((a-b)/(a+b))^(1/2)*(1/(1+cos
(d*x+c)))^(1/2)/a^3/(b+a*cos(d*x+c))/(a-b)/sin(d*x+c)^3

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.80, size = 804, normalized size = 2.47 \begin {gather*} \frac {6 \, {\left (A a^{4} b + 3 \, B a^{3} b^{2} - 4 \, A a^{2} b^{3} + {\left (A a^{5} - A a^{3} b^{2}\right )} \cos \left (d x + c\right )\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 ) - {\left (\sqrt {2} {\left (3 i \, A a^{5} - 15 i \, B a^{4} b + 16 i \, A a^{3} b^{2} + 12 i \, B a^{2} b^{3} - 16 i \, A a b^{4}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A a^{4} b - 15 i \, B a^{3} b^{2} + 16 i \, A a^{2} b^{3} + 12 i \, B a b^{4} - 16 i \, A b^{5}\right )}\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 ) - {\left (\sqrt {2} {\left (-3 i \, A a^{5} + 15 i \, B a^{4} b - 16 i \, A a^{3} b^{2} - 12 i \, B a^{2} b^{3} + 16 i \, A a b^{4}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A a^{4} b + 15 i \, B a^{3} b^{2} - 16 i \, A a^{2} b^{3} - 12 i \, B a b^{4} + 16 i \, A b^{5}\right )}\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 \, {\left (\sqrt {2} {\left (3 i \, B a^{5} - 5 i \, A a^{4} b - 6 i \, B a^{3} b^{2} + 8 i \, A a^{2} b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, B a^{4} b - 5 i \, A a^{3} b^{2} - 6 i \, B a^{2} b^{3} + 8 i \, A a b^{4}\right )}\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 \, {\left (\sqrt {2} {\left (-3 i \, B a^{5} + 5 i \, A a^{4} b + 6 i \, B a^{3} b^{2} - 8 i \, A a^{2} b^{3}\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, B a^{4} b + 5 i \, A a^{3} b^{2} + 6 i \, B a^{2} b^{3} - 8 i \, A a b^{4}\right )}\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 )}{9 \, {\left ({\left (a^{7} - a^{5} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - a^{4} b^{3}\right )} d\right )}} \end {gather*}

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]

1/9*(6*(A*a^4*b + 3*B*a^3*b^2 - 4*A*a^2*b^3 + (A*a^5 - A*a^3*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(
d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - (sqrt(2)*(3*I*A*a^5 - 15*I*B*a^4*b + 16*I*A*a^3*b^2 + 12*I*B*a^2*b
^3 - 16*I*A*a*b^4)*cos(d*x + c) + sqrt(2)*(3*I*A*a^4*b - 15*I*B*a^3*b^2 + 16*I*A*a^2*b^3 + 12*I*B*a*b^4 - 16*I
*A*b^5))*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) - (sqrt(2)*(-3*I*A*a^5 + 15*I*B*a^4*b - 16*I*A*a^3*b^2 - 12*I*B*a^2*b^3 + 16
*I*A*a*b^4)*cos(d*x + c) + sqrt(2)*(-3*I*A*a^4*b + 15*I*B*a^3*b^2 - 16*I*A*a^2*b^3 - 12*I*B*a*b^4 + 16*I*A*b^5
))*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*B*a^5 - 5*I*A*a^4*b - 6*I*B*a^3*b^2 + 8*I*A*a^2*b^3)*cos(d*x + c
) + sqrt(2)*(3*I*B*a^4*b - 5*I*A*a^3*b^2 - 6*I*B*a^2*b^3 + 8*I*A*a*b^4))*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*B*a^5 + 5*I*A*a^4*b + 6*I*B*a^3*
b^2 - 8*I*A*a^2*b^3)*cos(d*x + c) + sqrt(2)*(-3*I*B*a^4*b + 5*I*A*a^3*b^2 + 6*I*B*a^2*b^3 - 8*I*A*a*b^4))*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)))/((a^7 - a^5*b^2)*
d*cos(d*x + c) + (a^6*b - a^4*b^3)*d)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3435 deep

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________