3.1442 \(\int \frac {(A+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=348 \[ \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 (a (A-C)+2 A b) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d \sqrt {a+b} \sqrt {\sec (c+d x)}}-\frac {2 \left (2 A b^2-a^2 (A-C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^3 d \sqrt {a+b} \sqrt {\sec (c+d x)}} \]

[Out]

2*(A*b^2+C*a^2)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-2*(2*A*b^2-a^2*(A-C))*csc(d*x
+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*cos(d*x+c)^(1/2)*(a*(1
-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^3/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)-2*(2*A*b+a*(A-C))*
csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),((-a-b)/(a-b))^(1/2))*cos(d*x+c)^(1/2
)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d/(a+b)^(1/2)/sec(d*x+c)^(1/2)

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Rubi [A]  time = 0.82, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4221, 3056, 2998, 2816, 2994} \[ \frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (2 A b^2-a^2 (A-C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^3 d \sqrt {a+b} \sqrt {\sec (c+d x)}}-\frac {2 (a (A-C)+2 A b) \sqrt {\cos (c+d x)} \csc (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d \sqrt {a+b} \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*A*b^2 - a^2*(A - C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a
+ b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x])
)/(a - b)])/(a^3*Sqrt[a + b]*d*Sqrt[Sec[c + d*x]]) - (2*(2*A*b + a*(A - C))*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*El
lipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Se
c[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*Sqrt[a + b]*d*Sqrt[Sec[c + d*x]]) + (2*(A*b^2
+ a^2*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]])

Rule 2816

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rule 2994

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*A*(c - d)*Tan[e + f*x]*Rt[(c + d)/b, 2]*Sqrt[(c*(1 + Csc[e + f*x]))/(c
- d)]*Sqrt[(c*(1 - Csc[e + f*x]))/(c + d)]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[
(c + d)/b, 2])], -((c + d)/(c - d))])/(f*b*c^2), x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] &&
EqQ[A, B] && PosQ[(c + d)/b]

Rule 2998

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*s
in[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A - B)/(a - b), Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e
+ f*x]]), x], x] - Dist[(A*b - a*B)/(a - b), Int[(1 + Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin
[e + f*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] && NeQ[A, B]

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + 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] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 4221

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps

\begin {align*} \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx\\ &=\frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} \left (-2 A b^2+a^2 (A-C)\right )-\frac {1}{2} a b (A+C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {\left ((a-b) (2 A b+a (A-C)) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx}{a \left (a^2-b^2\right )}-\frac {\left (\left (2 A b^2-a^2 (A-C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {2 \left (2 A b^2-a^2 (A-C)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^3 \sqrt {a+b} d \sqrt {\sec (c+d x)}}-\frac {2 (2 A b+a (A-C)) \sqrt {\cos (c+d x)} \csc (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 \left (A b^2+a^2 C\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 17.73, size = 456, normalized size = 1.31 \[ \frac {\sqrt {\sec (c+d x)} \sqrt {a+b \cos (c+d x)} \left (\frac {2 \left (a^2 A-a^2 C-2 A b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right )}+\frac {2 \left (a^2 C \sin (c+d x)+A b^2 \sin (c+d x)\right )}{a \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )}{d}+\frac {2 \sqrt {2} \sqrt {\frac {\cos (c+d x)}{(\cos (c+d x)+1)^2}} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (\left (a^2 (C-A)+2 A b^2\right ) \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))-(a+b) \sec (c+d x) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a+b \cos (c+d x))}{a+b}} \left (\left (a^2 (A-C)-2 A b^2\right ) E\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+a (a (C-A)+2 A b) F\left (\sin ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )\right )}{a^2 d \left (a^2-b^2\right ) \sqrt {\frac {1}{\cos (c+d x)+1}} \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \sqrt {a+b \cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(a^2*A - 2*A*b^2 - a^2*C)*Sin[c + d*x])/(a^2*(a^2 - b^2)) + (
2*(A*b^2*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(a*(a^2 - b^2)*(a + b*Cos[c + d*x]))))/d + (2*Sqrt[2]*Sqrt[Cos[c
+ d*x]/(1 + Cos[c + d*x])^2]*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(-(
(a + b)*((-2*A*b^2 + a^2*(A - C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + a*(2*A*b + a*(-A + C
))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((a + b
*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x]) + (2*A*b^2 + a^2*(-A + C))*Cos[c + d*x]*(a + b*Cos[c
+ d*x])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2]))/(a^2*(a^2 - b^2)*d*Sqrt[(1 + Cos[c + d*x])^(-1)]*Sqrt[a + b*Cos
[c + d*x]]*(Sec[(c + d*x)/2]^2)^(3/2))

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fricas [F]  time = 1.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.66, size = 2287, normalized size = 6.57 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*(C*cos(d*x+c)^2*a^3-A*a^3-C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))
/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*a^2*b-A*(cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+
b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^3+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*b^3+A*sin(d*x+c)*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d
*x+c),(-(a-b)/(a+b))^(1/2))*a^3+A*a*b^2-A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*
x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b+2*A*sin(d*x
+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+c
os(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2-2*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2
-A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellip
ticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b-2*A*cos(d*x+c)^2*b^3+A*cos(d*x+c)^2*a^2*b+A*cos(d*
x+c)^2*a*b^2-A*cos(d*x+c)*a^2*b-2*A*cos(d*x+c)*a*b^2+A*cos(d*x+c)*a^3-2*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/
2))*a*b^2-A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipti
cE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b+2*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a
+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2+2*
A*cos(d*x+c)*b^3-C*cos(d*x+c)^2*a^2*b+C*cos(d*x+c)*a^2*b-A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b-C*cos
(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((
-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3+A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3-A
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin
(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*b^3-C*cos(d*x+c)*a^3-
C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos
(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b+C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c
))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b+C*sin(d*x+c)*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d
*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*a^3+C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1
+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*a^2*b+C*sin(d*
x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))
/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3-C*sin(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^3)*cos(d*x+c)*(1/cos(d*x+c))
^(3/2)/(a+b*cos(d*x+c))^(1/2)/sin(d*x+c)/a^2/(a-b)/(a+b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + b*cos(c + d*x))^(3/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(3/2))/(a + b*cos(c + d*x))^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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