3.1436 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^{\frac{5}{2}}(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=323 \[ \frac{2 \sqrt{a+b} (a (A+3 C)+2 A b) \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{4 A b (a-b) \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^3 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}{3 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.679344, antiderivative size = 323, normalized size of antiderivative = 1., 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} \[ \frac{2 \sqrt{a+b} (a (A+3 C)+2 A b) \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{4 A b (a-b) \sqrt{a+b} \sqrt{\cos (c+d x)} \csc (c+d x) \sqrt{\frac{a (1-\sec (c+d x))}{a+b}} \sqrt{\frac{a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \cos (c+d x)}}{\sqrt{a+b} \sqrt{\cos (c+d x)}}\right )|-\frac{a+b}{a-b}\right )}{3 a^3 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}}{3 a d} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

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

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]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{a+b \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-A b+\frac{1}{2} a (A+3 C) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \cos (c+d x)}} \, dx}{3 a}\\ &=\frac{2 A \sqrt{a+b \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}-\frac{\left (2 A b \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}{3 a}+\frac{\left ((2 A b+a (A+3 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}{3 a}\\ &=-\frac{4 A (a-b) b \sqrt{a+b} \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}}}{3 a^3 d \sqrt{\sec (c+d x)}}+\frac{2 \sqrt{a+b} (2 A b+a (A+3 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}}}{3 a^2 d \sqrt{\sec (c+d x)}}+\frac{2 A \sqrt{a+b \cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 13.2887, size = 303, normalized size = 0.94 \[ \frac{2 \left (\frac{2 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \left (a (a (A+3 C)-2 A b) \sqrt{\frac{1}{\sec (c+d x)+1}} \sqrt{\frac{a \sec (c+d x)+b}{(a+b) (\sec (c+d x)+1)}} F\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )+A b \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a+b \cos (c+d x))+2 A b (a+b) \sqrt{\frac{1}{\sec (c+d x)+1}} \sqrt{\frac{a \sec (c+d x)+b}{(a+b) (\sec (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{\sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )}}+A \sqrt{\sec (c+d x)} (a+b \cos (c+d x)) (a \tan (c+d x)-2 b \sin (c+d x))\right )}{3 a^2 d \sqrt{a+b \cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2))/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.243, size = 1102, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x)

[Out]

-2/3/d/a^2*(2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(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)^2*a*b+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(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)^2*b^2+A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2
)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)^2*a^2-2*A*(cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/
(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)^2*a*b+3*C*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+
b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2+2*A*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(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*b+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*co
s(d*x+c))/(1+cos(d*x+c)))^(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^2+A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d
*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*a^2-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+
b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*sin(d*x+c
)*cos(d*x+c)*a*b+3*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(
d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2+A*cos(d*x+c)^3*a*b-2*A*cos(d*x+c
)^3*b^2+A*cos(d*x+c)^2*a^2-2*A*cos(d*x+c)^2*a*b+2*A*cos(d*x+c)^2*b^2+A*cos(d*x+c)*a*b-A*a^2)*cos(d*x+c)*(1/cos
(d*x+c))^(5/2)/(a+b*cos(d*x+c))^(1/2)/sin(d*x+c)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{b \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/sqrt(b*cos(d*x + c) + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{b \cos \left (d x + c\right ) + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/sqrt(b*cos(d*x + c) + a), x)

________________________________________________________________________________________

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((A+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2)/(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{b \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^(5/2)/sqrt(b*cos(d*x + c) + a), x)