3.1324 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=346 \[ -\frac{\text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \left (-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)+12 a^3 b B-9 a b^3 B+15 A b^4\right )}{3 a^4 d \left (a^2-b^2\right )}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b (4 A-C)+2 a^3 B-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right )}+\frac{b \left (-a^2 b^2 (7 A-C)+5 a^3 b B-3 a^4 C-3 a b^3 B+5 A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2}+\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 (-(2 A-3 C))-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )} \]

[Out]

((5*A*b^3 + 2*a^3*B - 3*a*b^2*B - a^2*b*(4*A - C))*EllipticE[(c + d*x)/2, 2])/(a^3*(a^2 - b^2)*d) - ((15*A*b^4
+ 12*a^3*b*B - 9*a*b^3*B - a^2*b^2*(16*A - 3*C) - 2*a^4*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*a^4*(a^2 - b
^2)*d) + (b*(5*A*b^4 + 5*a^3*b*B - 3*a*b^3*B - a^2*b^2*(7*A - C) - 3*a^4*C)*EllipticPi[(2*a)/(a + b), (c + d*x
)/2, 2])/(a^4*(a - b)*(a + b)^2*d) - ((5*A*b^2 - 3*a*b*B - a^2*(2*A - 3*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
3*a^2*(a^2 - b^2)*d) + ((A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(b + a*Cos[c
+ d*x]))

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Rubi [A]  time = 1.21453, antiderivative size = 346, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4112, 3047, 3049, 3059, 2639, 3002, 2641, 2805} \[ -\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b^2 (16 A-3 C)-2 a^4 (A+3 C)+12 a^3 b B-9 a b^3 B+15 A b^4\right )}{3 a^4 d \left (a^2-b^2\right )}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (-a^2 b (4 A-C)+2 a^3 B-3 a b^2 B+5 A b^3\right )}{a^3 d \left (a^2-b^2\right )}+\frac{b \left (-a^2 b^2 (7 A-C)+5 a^3 b B-3 a^4 C-3 a b^3 B+5 A b^4\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2}+\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)} \left (a^2 (-(2 A-3 C))-3 a b B+5 A b^2\right )}{3 a^2 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*A*b^3 + 2*a^3*B - 3*a*b^2*B - a^2*b*(4*A - C))*EllipticE[(c + d*x)/2, 2])/(a^3*(a^2 - b^2)*d) - ((15*A*b^4
+ 12*a^3*b*B - 9*a*b^3*B - a^2*b^2*(16*A - 3*C) - 2*a^4*(A + 3*C))*EllipticF[(c + d*x)/2, 2])/(3*a^4*(a^2 - b
^2)*d) + (b*(5*A*b^4 + 5*a^3*b*B - 3*a*b^3*B - a^2*b^2*(7*A - C) - 3*a^4*C)*EllipticPi[(2*a)/(a + b), (c + d*x
)/2, 2])/(a^4*(a - b)*(a + b)^2*d) - ((5*A*b^2 - 3*a*b*B - a^2*(2*A - 3*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
3*a^2*(a^2 - b^2)*d) + ((A*b^2 - a*(b*B - a*C))*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(a*(a^2 - b^2)*d*(b + a*Cos[c
+ d*x]))

Rule 4112

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rule 3047

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 3049

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 3059

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]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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]

Rubi steps

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

Mathematica [A]  time = 3.68558, size = 339, normalized size = 0.98 \[ \frac{4 \sin (c+d x) \sqrt{\cos (c+d x)} \left (\frac{3 b \left (a (a C-b B)+A b^2\right )}{\left (b^2-a^2\right ) (a \cos (c+d x)+b)}+2 A\right )-\frac{\frac{8 \left (a^2 (A+3 C)-3 a b B+2 A b^2\right ) \left ((a+b) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )-b \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}-\frac{6 \sin (c+d x) \left (a^2 b (C-4 A)+2 a^3 B-3 a b^2 B+5 A b^3\right ) \left (-2 b (a+b) \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \Pi \left (-\frac{a}{b};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a^2 b \sqrt{\sin ^2(c+d x)}}+\frac{2 \left (-a^2 b (8 A+3 C)+6 a^3 B-3 a b^2 B+5 A b^3\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}}{(b-a) (a+b)}}{12 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[Cos[c + d*x]]*(2*A + (3*b*(A*b^2 + a*(-(b*B) + a*C)))/((-a^2 + b^2)*(b + a*Cos[c + d*x])))*Sin[c + d*x
] - ((2*(5*A*b^3 + 6*a^3*B - 3*a*b^2*B - a^2*b*(8*A + 3*C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b)
+ (8*(2*A*b^2 - 3*a*b*B + a^2*(A + 3*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c
+ d*x)/2, 2]))/(a + b) - (6*(5*A*b^3 + 2*a^3*B - 3*a*b^2*B + a^2*b*(-4*A + C))*(2*a*b*EllipticE[ArcSin[Sqrt[Co
s[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)*(a + b)))/(12*a^2*d)

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Maple [B]  time = 9.059, size = 1123, normalized size = 3.3 \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)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)

[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*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^4+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-2
*A*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-6*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+3*a^2*C*(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)))/(-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*b^2-B*
a*b+C*a^2)/a^4*(a^2/b/(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*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)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*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)*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)*Ellipt
icPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))+2/a^3*b*(4*A*b^2-3*B*a*b+2*C*a^2)/(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)))/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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + 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 )}^{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)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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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)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + 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 )}^{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)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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