3.6.90 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\) [590]

Optimal. Leaf size=338 \[ -\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4 A b+10 a^2 A b^3-A b^5+a^5 B-10 a^3 b^2 B-3 a b^4 B\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 (a-b)^2 b (a+b)^3 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \]

[Out]

-1/4*(5*A*a^2*b+A*b^3-B*a^3-5*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/b/(a^2-b^2)^2/d-1/4*(7*A*a^2*b-A*b^3-3*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)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^2)^2/d+1/4*(3*A*a^4*b+10*A*a^2*b^3-A*b^5+B*a^5-10*
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^2/(a-b)^2/b/(a+b)^3/d-1/2*(A*b-B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)/(a^2-b^2)/d/(b+a*cos(d*x+c))^2+1/4*(
5*A*a^2*b+A*b^3-B*a^3-5*B*a*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/b/(a^2-b^2)^2/d/(b+a*cos(d*x+c))

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Rubi [A]
time = 0.66, antiderivative size = 338, 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, 3078, 3134, 3138, 2719, 3081, 2720, 2884} \begin {gather*} -\frac {(A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 d \left (a^2-b^2\right )^2}-\frac {\left (a^3 (-B)+5 a^2 A b-5 a b^2 B+A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b d \left (a^2-b^2\right )^2}+\frac {\left (a^3 (-B)+5 a^2 A b-5 a b^2 B+A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{4 b d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac {\left (a^5 B+3 a^4 A b-10 a^3 b^2 B+10 a^2 A b^3-3 a b^4 B-A b^5\right ) \Pi \left (\frac {2 a}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 b d (a-b)^2 (a+b)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((5*a^2*A*b + A*b^3 - a^3*B - 5*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(a*b*(a^2 - b^2)^2*d) - ((7*a^2*A*b -
A*b^3 - 3*a^3*B - 3*a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(4*a^2*(a^2 - b^2)^2*d) + ((3*a^4*A*b + 10*a^2*A*b^3
- A*b^5 + a^5*B - 10*a^3*b^2*B - 3*a*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(4*a^2*(a - b)^2*b*(a +
b)^3*d) - ((A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*(a^2 - b^2)*d*(b + a*Cos[c + d*x])^2) + ((5*a^2*A*
b + A*b^3 - a^3*B - 5*a*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(4*b*(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 3078

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*a - A*b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e
+ f*x])^n/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n - 1)*Simp[c*(a*A - b*B)*(m + 1) + d*n*(A*b - a*B) + (d*(a*A - b*B)*(m + 1) - c*(A*b - a*B)*
(m + 2))*Sin[e + f*x] - d*(A*b - a*B)*(m + n + 2)*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] && LtQ[m, -1] && GtQ[n, 0]

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 3134

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[(-(A*b^2 - a*b*B + a^2*C))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*
(b*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(
b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 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 {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx &=\int \frac {\sqrt {\cos (c+d x)} (B+A \cos (c+d x))}{(b+a \cos (c+d x))^3} \, dx\\ &=-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} (-A b+a B)+2 (a A-b B) \cos (c+d x)-\frac {1}{2} (A b-a B) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\int \frac {\frac {1}{4} \left (-3 a^2 A b-3 A b^3-a^3 B+7 a b^2 B\right )+b \left (3 a A b-a^2 B-2 b^2 B\right ) \cos (c+d x)+\frac {1}{4} \left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 b \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{4} a \left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right )-\frac {1}{4} b \left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{2 a b \left (a^2-b^2\right )^2}-\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx}{8 a b \left (a^2-b^2\right )^2}\\ &=-\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{8 a^2 \left (a^2-b^2\right )^2}+\frac {\left (3 a^4 A b+10 a^2 A b^3-A b^5+a^5 B-10 a^3 b^2 B-3 a b^4 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{8 a^2 b \left (a^2-b^2\right )^2}\\ &=-\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4 A b+10 a^2 A b^3-A b^5+a^5 B-10 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^2 (a-b)^2 b (a+b)^3 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 14.86, size = 364, normalized size = 1.08 \begin {gather*} \frac {\frac {2 \sqrt {\cos (c+d x)} \left (b \left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right )-a \left (-5 a^2 A b-A b^3+a^3 B+5 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 (a^2 A b+5 A b^3+3 a^3 B-9 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 b \left (-3 a A b+a^2 B+2 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 (a+b)}+\frac {\left (-5 a^2 A b-A b^3+a^3 B+5 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)^2 (a+b)^2}}{8 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[Cos[c + d*x]]*(b*(3*a^2*A*b + 3*A*b^3 + a^3*B - 7*a*b^2*B) - a*(-5*a^2*A*b - A*b^3 + a^3*B + 5*a*b^2*
B)*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (((a^2*A*b + 5*A*b^3 + 3*a^3*B - 9*a*b
^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*b*(-3*a*A*b + a^2*B + 2*b^2*B)*((a + b)*Elliptic
F[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a*(a + b)) + ((-5*a^2*A*b - A*b^3 + a^3*B +
5*a*b^2*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*b*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1871\) vs. \(2(402)=804\).
time = 10.99, size = 1872, normalized size = 5.54

method	result	size
default	\(\text {Expression too large to display}\)	\(1872\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(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*(-2*A*b+B*a)/a^2*(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)*E
llipticF(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)*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(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))-
2*A/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*(A*b-B*a)/a^2*(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)*EllipticF(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(c
os(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+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))-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

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

[Out]

Timed out

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

[Out]

Timed out

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

[Out]

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

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

[Out]

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

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