3.6.88 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [588]
Optimal. Leaf size=367 \[ \frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \]
[Out]
1/4*(8*A*a^4-29*A*a^2*b^2+15*A*b^4+9*B*a^3*b-3*B*a*b^3)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/(a^2-b^2)^2/d-1/4*(24*A*a^4*b-33*A*a^2*b^3+15*A*b^5-8*B*a^5+5*B*a^3*b^2-3*
B*a*b^4)*(cos(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^4/(a^2-b^2)^2
/d+1/4*b*(35*A*a^4*b-38*A*a^2*b^3+15*A*b^5-15*B*a^5+6*B*a^3*b^2-3*B*a*b^4)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/
2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^4/(a-b)^2/(a+b)^3/d+1/2*b*(A*b-B*a)*cos(d*x+c)
^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))^2+1/4*b*(11*A*a^2*b-5*A*b^3-7*B*a^3+B*a*b^2)*sin(d*x+c)*cos(d
*x+c)^(1/2)/a^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))
________________________________________________________________________________________
Rubi [A]
time = 0.71, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3033, 3068,
3126, 3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {\left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 d \left (a^2-b^2\right )^2}-\frac {\left (-8 a^5 B+24 a^4 A b+5 a^3 b^2 B-33 a^2 A b^3-3 a b^4 B+15 A b^5\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d \left (a^2-b^2\right )^2}+\frac {b \left (-15 a^5 B+35 a^4 A b+6 a^3 b^2 B-38 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 d (a-b)^2 (a+b)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3,x]
[Out]
((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*EllipticE[(c + d*x)/2, 2])/(4*a^3*(a^2 - b^2)^2*d
) - ((24*a^4*A*b - 33*a^2*A*b^3 + 15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B - 3*a*b^4*B)*EllipticF[(c + d*x)/2, 2])/(4*
a^4*(a^2 - b^2)^2*d) + (b*(35*a^4*A*b - 38*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 6*a^3*b^2*B - 3*a*b^4*B)*Elliptic
Pi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^4*(a - b)^2*(a + b)^3*d) + (b*(A*b - a*B)*Cos[c + d*x]^(3/2)*Sin[c + d
*x])/(2*a*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + (b*(11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(4*a^2*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))
Rule 2719
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]
Rule 2720
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]
Rule 2884
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 3033
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^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 3068
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1
)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Si
n[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c -
(A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*
d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 1] && LtQ[n, -1]
Rule 3081
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Rule 3126
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)
*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) +
(c*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*
c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n +
1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Rule 3138
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]
Rubi steps
\begin {align*} \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx &=\int \frac {\cos ^{\frac {5}{2}}(c+d x) (B+A \cos (c+d x))}{(b+a \cos (c+d x))^3} \, dx\\ &=\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} b (A b-a B)-2 a (A b-a B) \cos (c+d x)+\frac {1}{2} \left (4 a^2 A-5 A b^2+a b B\right ) \cos ^2(c+d x)\right )}{(b+a \cos (c+d x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right )-a \left (4 a^2 A b-A b^3-2 a^3 B-a b^2 B\right ) \cos (c+d x)+\frac {1}{4} \left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\int \frac {-\frac {1}{4} a b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right )+\frac {1}{4} \left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^3 \left (a^2-b^2\right )^2}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a^3 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^4 \left (a^2-b^2\right )^2}+\frac {\left (b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^4 \left (a^2-b^2\right )^2}\\ &=\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 15.07, size = 390, normalized size = 1.06 \begin {gather*} \frac {-\frac {2 b \sqrt {\cos (c+d x)} \left (b \left (-11 a^2 A b+5 A b^3+7 a^3 B-a b^2 B\right )+a \left (-13 a^2 A b+7 A b^3+9 a^3 B-3 a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {\left (8 a^4 A-7 a^2 A b^2+5 A b^4-5 a^3 b B-a b^3 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 \left (-4 a^2 A b+A b^3+2 a^3 B+a b^2 B\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-b \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \left (-2 a b E\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) F\left (\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac {a}{b};\left .\text {ArcSin}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 a^2 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3,x]
[Out]
((-2*b*Sqrt[Cos[c + d*x]]*(b*(-11*a^2*A*b + 5*A*b^3 + 7*a^3*B - a*b^2*B) + a*(-13*a^2*A*b + 7*A*b^3 + 9*a^3*B
- 3*a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (((8*a^4*A - 7*a^2*A*b^2 + 5
*A*b^4 - 5*a^3*b*B - a*b^3*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(-4*a^2*A*b + A*b^3 + 2*
a^3*B + a*b^2*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) +
((8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1]
+ 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c
+ d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(8*a^2*d)
________________________________________________________________________________________
Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1999\) vs.
\(2(431)=862\).
time = 12.32, size = 2000, normalized size = 5.45
| | |
| method |
result |
size |
| | |
| default |
\(\text {Expression too large to display}\) |
\(2000\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/a^4/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))*b+A*a*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*a*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*b^2/a^4*(4*A*b-3
*B*a)*(1/b*a^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*
x+1/2*c)^2*a-a+b)-1/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2/b*a/(a^2-b^2)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))-1/2/b*a/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a
*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(co
s(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-6*b/a^3*(2*A*b-B*a)/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-
b),2^(1/2))+2*b^3*(A*b-B*a)/a^4*(1/2/b*a^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1
/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*sin
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2*a-a+b)-3/8/(a+b)/(a^2-b^2)/b^2*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-1/4/(a+b)/(a^2-b^2)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*
c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a+7/8
/(a+b)/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3/8*a^3/b^2/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d
*x+1/2*c),2^(1/2))-9/8*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3/8*a^3/b^2/(a^2-b^2)^2*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+9/8*a/(a^2-b^2)^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2
+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3/8/(a-b)
/(a+b)/(a^2-b^2)/b^2/(a^2-a*b)*a^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/4/(a-b)/(a+b)/(a^2
-b^2)/(a^2-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-15/8/(a-b)/(a+b)/(a^2-b^2)*b^2/(a^2
-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)
^(1/2)/d
________________________________________________________________________________________
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((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [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((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))*cos(d*x+c)**(1/2)/(a+b*sec(d*x+c))**3,x)
[Out]
Integral((A + B*sec(c + d*x))*sqrt(cos(c + d*x))/(a + b*sec(c + d*x))**3, x)
________________________________________________________________________________________
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((A+B*sec(d*x+c))*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(b*sec(d*x + c) + a)^3, x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^3,x)
[Out]
int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x)))/(a + b/cos(c + d*x))^3, x)
________________________________________________________________________________________