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

Optimal. Leaf size=245 \[ \frac{(8 A+19 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(5 A+13 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(A+2 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 a d \sqrt{a \cos (c+d x)+a}}-\frac{(2 A+7 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

((8*A + 19*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(3/2)*d) - ((5*A + 13*C)*ArcTan[(S
qrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*
Cos[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((2*A + 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*
x])/(4*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((A + 2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d
*x]])

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Rubi [A]  time = 0.786682, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used = {3042, 2983, 2982, 2782, 205, 2774, 216} \[ \frac{(8 A+19 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac{(5 A+13 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac{(A+2 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 a d \sqrt{a \cos (c+d x)+a}}-\frac{(2 A+7 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{4 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*A + 19*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(3/2)*d) - ((5*A + 13*C)*ArcTan[(S
qrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*
Cos[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((2*A + 7*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*
x])/(4*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((A + 2*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d
*x]])

Rule 3042

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*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2983

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*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2982

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/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] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac{(A+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\cos ^{\frac{3}{2}}(c+d x) \left (-\frac{1}{2} a (A+5 C)+2 a (A+2 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{(A+2 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (3 a^2 (A+2 C)-a^2 (2 A+7 C) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a^3}\\ &=-\frac{(A+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(2 A+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a d \sqrt{a+a \cos (c+d x)}}+\frac{(A+2 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-\frac{1}{2} a^3 (2 A+7 C)+\frac{1}{2} a^3 (8 A+19 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a^4}\\ &=-\frac{(A+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(2 A+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a d \sqrt{a+a \cos (c+d x)}}+\frac{(A+2 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}-\frac{(5 A+13 C) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx}{4 a}+\frac{(8 A+19 C) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{8 a^2}\\ &=-\frac{(A+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(2 A+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a d \sqrt{a+a \cos (c+d x)}}+\frac{(A+2 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}+\frac{(5 A+13 C) \operatorname{Subst}\left (\int \frac{1}{2 a^2+a x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{2 d}-\frac{(8 A+19 C) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a}}} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 a^2 d}\\ &=\frac{(8 A+19 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac{(5 A+13 C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A+C) \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(2 A+7 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{4 a d \sqrt{a+a \cos (c+d x)}}+\frac{(A+2 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 2.40302, size = 370, normalized size = 1.51 \[ \frac{\cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (-2 \sqrt{\cos (c+d x)} \tan \left (\frac{1}{2} (c+d x)\right ) \sec \left (\frac{1}{2} (c+d x)\right ) (2 A+3 C \cos (c+d x)-C \cos (2 (c+d x))+6 C)+\frac{\sqrt{2} e^{\frac{1}{2} i (c+d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (2 i \sqrt{2} (5 A+13 C) \log \left (1+e^{i (c+d x)}\right )-i (8 A+19 C) \sinh ^{-1}\left (e^{i (c+d x)}\right )+8 i A \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-10 i \sqrt{2} A \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+8 A d x+19 i C \log \left (1+\sqrt{1+e^{2 i (c+d x)}}\right )-26 i \sqrt{2} C \log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )+19 C d x\right )}{\sqrt{1+e^{2 i (c+d x)}}}\right )}{4 d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*((Sqrt[2]*E^((I/2)*(c + d*x))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(8*A*d*x + 1
9*C*d*x - I*(8*A + 19*C)*ArcSinh[E^(I*(c + d*x))] + (2*I)*Sqrt[2]*(5*A + 13*C)*Log[1 + E^(I*(c + d*x))] + (8*I
)*A*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] + (19*I)*C*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]] - (10*I)*Sqrt[2]*
A*Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] - (26*I)*Sqrt[2]*C*Log[1 - E^(I*(c + d*x))
+ Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/Sqrt[1 + E^((2*I)*(c + d*x))] - 2*Sqrt[Cos[c + d*x]]*(2*A + 6*C + 3
*C*Cos[c + d*x] - C*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]*Tan[(c + d*x)/2]))/(4*d*(a*(1 + Cos[c + d*x]))^(3/2))

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Maple [B]  time = 0.16, size = 477, normalized size = 2. \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+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(3/2),x)

[Out]

1/4/d*cos(d*x+c)^(3/2)*(a*(1+cos(d*x+c)))^(1/2)*(-1+cos(d*x+c))^4*(2*A*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c))
)^(5/2)+2*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-2*A*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-2*
C*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+5*A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos
(d*x+c)^2-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+13*C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)*sin(d*x+c)*cos
(d*x+c)^2+5*C*cos(d*x+c)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+8*A*sin(d*x+c)*cos(d*x+c)^2*arctan(sin(d*x+c)*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d*x+c))+19*C*sin(d*x+c)*cos(d*x+c)^2*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)/cos(d*x+c))+4*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-7*C*cos(d*x+c)^2*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2))/sin(d*x+c)^9/(cos(d*x+c)/(1+cos(d*x+c)))^(7/2)/a^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 88.8205, size = 678, normalized size = 2.77 \begin{align*} \frac{\sqrt{2}{\left ({\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right ) + 5 \, A + 13 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) +{\left (2 \, C \cos \left (d x + c\right )^{2} - 3 \, C \cos \left (d x + c\right ) - 2 \, A - 7 \, C\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) -{\left ({\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 19 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*((5*A + 13*C)*cos(d*x + c)^2 + 2*(5*A + 13*C)*cos(d*x + c) + 5*A + 13*C)*sqrt(a)*arctan(sqrt(2)*s
qrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + (2*C*cos(d*x + c)^2 - 3*C*cos(d*x + c) -
2*A - 7*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - ((8*A + 19*C)*cos(d*x + c)^2 + 2*(8*A +
19*C)*cos(d*x + c) + 8*A + 19*C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x +
c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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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+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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