[next] [prev] [prev-tail] [tail] [up]

3.406 \(\int \frac {(a+b \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=419 \[ \frac {2 \sqrt {a+b} \left (a^2 (A-3 B)-a (4 A b-6 b B)+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a d}+\frac {2 (a-b) \sqrt {a+b} (3 a B+4 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a d}+\frac {2 a A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b B \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{d} \]

[Out]

2/3*a*A*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(3/2)+2/3*(a-b)*(4*A*b+3*B*a)*cot(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))*(a+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/
2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/d+2/3*(3*A*b^2+a^2*(A-3*B)-a*(4*A*b-6*B*b))*cot(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))*(a+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a
*(1+sec(d*x+c))/(a-b))^(1/2)/a/d-2*b*B*cot(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/
2),(a+b)/b,((-a-b)/(a-b))^(1/2))*(a+b)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.86, antiderivative size = 419, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2989, 3053, 2809, 2998, 2816, 2994} \[ \frac {2 \sqrt {a+b} \left (a^2 (A-3 B)-a (4 A b-6 b B)+3 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a d}+\frac {2 (a-b) \sqrt {a+b} (3 a B+4 A b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{3 a d}+\frac {2 a A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 b B \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac {a+b}{b};\sin ^{-1}\left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]

[Out]

(2*(a - b)*Sqrt[a + b]*(4*A*b + 3*a*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqr
t[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*d) + (2*Sqrt[a + b]*(3*A*b^2 + a^2*(A - 3*B) - a*(4*A*b - 6*b*B))*Cot[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*d) - (2*b*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/b, 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)])/d + (2*a*A*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d
*x]^(3/2))

Rule 2809

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(2*b*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)]*EllipticP
i[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/(Sqrt[b*Sin[e + f*x]]*Rt[(c + d)/b, 2])], -((c + d)/(c - d))])/(d
*f), x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/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 2989

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*Sin[
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 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 3053

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_.) + (b_.)*sin[(e_.) + (f_.
)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[C/b^2, Int[Sqrt[a + b*Sin[e + f
*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[1/b^2, Int[(A*b^2 - a^2*C + b*(b*B - 2*a*C)*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, 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 \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {5}{2}}(c+d x)} \, dx &=\frac {2 a A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {1}{2} a (4 A b+3 a B)+\frac {1}{2} \left (a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)+\frac {3}{2} b^2 B \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 a A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {1}{2} a (4 A b+3 a B)+\frac {1}{2} \left (a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx+\left (b^2 B\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}} \, dx\\ &=-\frac {2 b \sqrt {a+b} B \cot (c+d x) \Pi \left (\frac {a+b}{b};\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}}}{d}+\frac {2 a A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {1}{3} (a (4 A b+3 a B)) \int \frac {1+\cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}} \, dx+\frac {1}{3} \left (3 A b^2+a^2 (A-3 B)-a (4 A b-6 b B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \, dx\\ &=\frac {2 (a-b) \sqrt {a+b} (4 A b+3 a B) \cot (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 d}+\frac {2 \sqrt {a+b} \left (3 A b^2+a^2 (A-3 B)-a (4 A b-6 b B)\right ) \cot (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 d}-\frac {2 b \sqrt {a+b} B \cot (c+d x) \Pi \left (\frac {a+b}{b};\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}}}{d}+\frac {2 a A \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.36, size = 1236, normalized size = 2.95 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]

[Out]

((-4*a*(a^2*A - A*b^2 + 3*a*b*B)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[
(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a +
 b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d
*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(-4*a*A*b - 3*a^2*B + 3*b^2*B)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a +
b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc
[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c
 + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a +
 b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Cs
c[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a +
b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(-4*A*b^2 - 3*a*b*B)*((I*Cos[(c +
 d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b)]*S
ec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x])*Sec[c + d*x])/(a + b)]) + (2*
a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((
a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d
*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x
]]) - (a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt
[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x]
)*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c
 + d*x]])))/b + (Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*Sqrt[Cos[c + d*x]])))/(3*d) + (Sqrt[Cos[c + d*x]]*S
qrt[a + b*Cos[c + d*x]]*((2*Sec[c + d*x]*(4*A*b*Sin[c + d*x] + 3*a*B*Sin[c + d*x]))/3 + (2*a*A*Sec[c + d*x]*Ta
n[c + d*x])/3))/d

________________________________________________________________________________________

fricas [F]  time = 2.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b \cos \left (d x + c\right )^{2} + A a + {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/2)/cos(d*x + c)^(5/2), x)

________________________________________________________________________________________

maple [B]  time = 0.40, size = 2318, normalized size = 5.53 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x)

[Out]

-2/3/d*(3*B*cos(d*x+c)^2*a^2+3*A*sin(d*x+c)*cos(d*x+c)^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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2+6*B*sin(d*x+c)*cos(d
*x+c)^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)*EllipticPi((-1+cos(d*x
+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*b^2-3*B*sin(d*x+c)*cos(d*x+c)^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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2+6*B*
EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*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)*cos(d*x+c)^2*a*b-3*B*cos(d*x+c)*a^2+4*A*cos(d*x+c)^3*b^2-4*A*cos(d*x+c
)^2*b^2-a^2*A-4*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)^2*a*b+A*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)*sin(
d*x+c)*cos(d*x+c)^2*a^2-4*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)*El
lipticE((-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+3*B*EllipticF((-1+cos(d*x
+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*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)*cos(d*x+c)^2*a^2-3*B*sin(d*x+c)*cos(d*x+c)^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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2+A*sin(d*x+c)*
cos(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^2+3*B*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*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)*cos(d*x+c)*a^2+A*cos(d
*x+c)^2*a^2-3*B*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)*Ell
ipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*a*b+6*B*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*b-4*A*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*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)*a*b+4*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)*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+A*cos(d*x+c)^3*a*b-3*B*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)^2*a*b+4*
A*sin(d*x+c)*cos(d*x+c)^2*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*b+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)*EllipticF((-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+4*A*cos(d*x+c)^2*a*b-5*A*cos(d*x+c)*a*b+3*B*cos(d*x+c)^3*a*b-3*B*cos(d*x+c)^2*a*b-4*A*EllipticE((-1+co
s(d*x+c))/sin(d*x+c),(-(a-b)/(a+b))^(1/2))*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)*b^2+6*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+b*cos(d*x+c))/(1+c
os(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*cos(d*x+c)*b^2-3*B*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^2-3*B*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)*b^2)
/(a+b*cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^(3/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/2)/cos(d*x + c)^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(3/2))/cos(c + d*x)^(5/2),x)

[Out]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^(3/2))/cos(c + d*x)^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c))/cos(d*x+c)**(5/2),x)

[Out]

Integral((A + B*cos(c + d*x))*(a + b*cos(c + d*x))**(3/2)/cos(c + d*x)**(5/2), x)

________________________________________________________________________________________

[next] [prev] [prev-tail] [front] [up]