3.6.81 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [581]

Optimal. Leaf size=305 \[ -\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b^2 \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 (a-b) (a+b)^2 d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.66, antiderivative size = 305, 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, 3128, 3138, 2719, 3081, 2720, 2884} \begin {gather*} \frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\left (2 a^2 A+3 a b B-5 A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a^2 d \left (a^2-b^2\right )}-\frac {\left (-2 a^3 B+4 a^2 A b+3 a b^2 B-5 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \left (a^2-b^2\right )}-\frac {b^2 \left (-5 a^3 B+7 a^2 A b+3 a b^2 B-5 A b^3\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2}+\frac {\left (2 a^4 A-12 a^3 b B+16 a^2 A b^2+9 a b^3 B-15 A b^4\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 d \left (a^2-b^2\right )} \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])^2,x]

[Out]

-(((4*a^2*A*b - 5*A*b^3 - 2*a^3*B + 3*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(a^3*(a^2 - b^2)*d)) + ((2*a^4*A + 1
6*a^2*A*b^2 - 15*A*b^4 - 12*a^3*b*B + 9*a*b^3*B)*EllipticF[(c + d*x)/2, 2])/(3*a^4*(a^2 - b^2)*d) - (b^2*(7*a^
2*A*b - 5*A*b^3 - 5*a^3*B + 3*a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a^4*(a - b)*(a + b)^2*d) +
((2*a^2*A - 5*A*b^2 + 3*a*b*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*d) + (b*(A*b - a*B)*Cos[c +
d*x]^(3/2)*Sin[c + d*x])/(a*(a^2 - b^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 3128

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

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 {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^{\frac {5}{2}}(c+d x) (B+A \cos (c+d x))}{(b+a \cos (c+d x))^2} \, dx\\ &=\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {3}{2} b (A b-a B)-a (A b-a B) \cos (c+d x)+\frac {1}{2} \left (2 a^2 A-5 A b^2+3 a b B\right ) \cos ^2(c+d x)\right )}{b+a \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {2 \int \frac {\frac {1}{4} b \left (2 a^2 A-5 A b^2+3 a b B\right )+\frac {1}{2} a \left (a^2 A+2 A b^2-3 a b B\right ) \cos (c+d x)-\frac {3}{4} \left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a^2 \left (a^2-b^2\right )}\\ &=\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac {2 \int \frac {-\frac {1}{4} a b \left (2 a^2 A-5 A b^2+3 a b B\right )-\frac {1}{4} \left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{3 a^3 \left (a^2-b^2\right )}-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}-\frac {\left (b^2 \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a^4 \left (a^2-b^2\right )}+\frac {\left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^4 \left (a^2-b^2\right )}\\ &=-\frac {\left (4 a^2 A b-5 A b^3-2 a^3 B+3 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (2 a^4 A+16 a^2 A b^2-15 A b^4-12 a^3 b B+9 a b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^4 \left (a^2-b^2\right ) d}-\frac {b^2 \left (7 a^2 A b-5 A b^3-5 a^3 B+3 a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^4 (a-b) (a+b)^2 d}+\frac {\left (2 a^2 A-5 A b^2+3 a b B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 13.38, size = 318, normalized size = 1.04 \begin {gather*} \frac {4 \sqrt {\cos (c+d x)} \left (2 A+\frac {3 b^2 (A b-a B)}{\left (-a^2+b^2\right ) (b+a \cos (c+d x))}\right ) \sin (c+d x)-\frac {\frac {2 \left (-8 a^2 A b+5 A b^3+6 a^3 B-3 a b^2 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {8 \left (a^2 A+2 A b^2-3 a b 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 {6 \left (-4 a^2 A b+5 A b^3+2 a^3 B-3 a b^2 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) (a+b)}}{12 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[Cos[c + d*x]]*(2*A + (3*b^2*(A*b - a*B))/((-a^2 + b^2)*(b + a*Cos[c + d*x])))*Sin[c + d*x] - ((2*(-8*a
^2*A*b + 5*A*b^3 + 6*a^3*B - 3*a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*(a^2*A + 2*A*b
^2 - 3*a*b*B)*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + (6*
(-4*a^2*A*b + 5*A*b^3 + 2*a^3*B - 3*a*b^2*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*E
llipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*S
in[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((-a + b)*(a + b)))/(12*a^2*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1058\) vs. \(2(375)=750\).
time = 6.83, size = 1059, normalized size = 3.47

method	result	size
default	\(\text {Expression too large to display}\)	\(1059\)

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))^2,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)*(2/3/a^4*(4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^4*a^2-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2+a^2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*A*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^
2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(
1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-6*B*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*b^2/a^3*(4*
A*b-3*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+s
in(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/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(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)*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)*Ellipt
icPi(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(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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [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))^2,x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*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)] 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))**2,x)

[Out]

Timed out

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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))^2,x, algorithm="giac")

[Out]

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

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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 )}^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))^2,x)

[Out]

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

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