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

Optimal. Leaf size=216 \[ \frac{(35 A-39 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{30 a^2 d}+\frac{(11 A-15 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(5 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}-\frac{(65 A-93 B) \sin (c+d x)}{15 a d \sqrt{a \cos (c+d x)+a}} \]

[Out]

((11*A - 15*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((A
- B)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((65*A - 93*B)*Sin[c + d*x])/(15*a*d*Sqr
t[a + a*Cos[c + d*x]]) - ((5*A - 9*B)*Cos[c + d*x]^2*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((35*A
- 39*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(30*a^2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.59499, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2977, 2983, 2968, 3023, 2751, 2649, 206} \[ \frac{(35 A-39 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{30 a^2 d}+\frac{(11 A-15 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac{(5 A-9 B) \sin (c+d x) \cos ^2(c+d x)}{10 a d \sqrt{a \cos (c+d x)+a}}-\frac{(65 A-93 B) \sin (c+d x)}{15 a d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((11*A - 15*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((A
- B)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((65*A - 93*B)*Sin[c + d*x])/(15*a*d*Sqr
t[a + a*Cos[c + d*x]]) - ((5*A - 9*B)*Cos[c + d*x]^2*Sin[c + d*x])/(10*a*d*Sqrt[a + a*Cos[c + d*x]]) + ((35*A
- 39*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(30*a^2*d)

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], 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] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2751

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

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\cos ^2(c+d x) \left (3 a (A-B)-\frac{1}{2} a (5 A-9 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (-a^2 (5 A-9 B)+\frac{1}{4} a^2 (35 A-39 B) \cos (c+d x)\right )}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{-a^2 (5 A-9 B) \cos (c+d x)+\frac{1}{4} a^2 (35 A-39 B) \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a^3}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{(35 A-39 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}+\frac{2 \int \frac{\frac{1}{8} a^3 (35 A-39 B)-\frac{1}{4} a^3 (65 A-93 B) \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(65 A-93 B) \sin (c+d x)}{15 a d \sqrt{a+a \cos (c+d x)}}-\frac{(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{(35 A-39 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}+\frac{(11 A-15 B) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{4 a}\\ &=\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(65 A-93 B) \sin (c+d x)}{15 a d \sqrt{a+a \cos (c+d x)}}-\frac{(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{(35 A-39 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}-\frac{(11 A-15 B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{2 a d}\\ &=\frac{(11 A-15 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac{(65 A-93 B) \sin (c+d x)}{15 a d \sqrt{a+a \cos (c+d x)}}-\frac{(5 A-9 B) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt{a+a \cos (c+d x)}}+\frac{(35 A-39 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{30 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.870648, size = 142, normalized size = 0.66 \[ \frac{\sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right ) (3 (20 A-39 B) \cos (c+d x)+(6 B-10 A) \cos (2 (c+d x))+85 A-3 B \cos (3 (c+d x))-141 B)-15 (11 A-15 B) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{15 d \left (\sin ^2\left (\frac{1}{2} (c+d x)\right )-1\right ) (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*(11*A - 15*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 + Cos[(c + d*x)/2]^3*(85*A - 141*B + 3*(20*A -
39*B)*Cos[c + d*x] + (-10*A + 6*B)*Cos[2*(c + d*x)] - 3*B*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])/(15*d*(a*(1 + C
os[c + d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

________________________________________________________________________________________

Maple [B]  time = 2.473, size = 407, normalized size = 1.9 \begin{align*}{\frac{1}{60\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -96\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+16\,\sqrt{2}\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( 5\,A+6\,B \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+5\,\sqrt{2} \left ( 8\,A\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-33\,A\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) a-48\,B\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+45\,B\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) a \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+165\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aA-225\,\sqrt{2}\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) aB-135\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+255\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{a}^{-{\frac{5}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-96*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*s
in(1/2*d*x+1/2*c)^6+16*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(5*A+6*B)*sin(1/2*d*x+1/2*c)^4+5*2^(1/2)
*(8*A*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-33*A*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1
/2)+a))*a-48*B*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+45*B*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2
*c)^2)^(1/2)+a))*a)*sin(1/2*d*x+1/2*c)^2+165*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)
^(1/2)+a))*a*A-225*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*B-135*A*a^(1/
2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+255*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2)/sin(1/
2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.79943, size = 601, normalized size = 2.78 \begin{align*} -\frac{15 \, \sqrt{2}{\left ({\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (11 \, A - 15 \, B\right )} \cos \left (d x + c\right ) + 11 \, A - 15 \, B\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left (12 \, B \cos \left (d x + c\right )^{3} + 4 \,{\left (5 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - 12 \,{\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right ) - 95 \, A + 147 \, B\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{120 \,{\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*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/120*(15*sqrt(2)*((11*A - 15*B)*cos(d*x + c)^2 + 2*(11*A - 15*B)*cos(d*x + c) + 11*A - 15*B)*sqrt(a)*log(-(a
*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x +
c)^2 + 2*cos(d*x + c) + 1)) - 4*(12*B*cos(d*x + c)^3 + 4*(5*A - 3*B)*cos(d*x + c)^2 - 12*(5*A - 9*B)*cos(d*x
+ c) - 95*A + 147*B)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2
*d)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.92885, size = 273, normalized size = 1.26 \begin{align*} -\frac{\frac{15 \, \sqrt{2}{\left (11 \, A - 15 \, B\right )} \log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac{3}{2}}} + \frac{{\left ({\left ({\left (\frac{15 \, \sqrt{2}{\left (A a^{3} - B a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{2}} + \frac{\sqrt{2}{\left (245 \, A a^{3} - 381 \, B a^{3}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{5 \, \sqrt{2}{\left (73 \, A a^{3} - 105 \, B a^{3}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{15 \, \sqrt{2}{\left (9 \, A a^{3} - 17 \, B a^{3}\right )}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{5}{2}}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt(2)*(11*A - 15*B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/a
^(3/2) + (((15*sqrt(2)*(A*a^3 - B*a^3)*tan(1/2*d*x + 1/2*c)^2/a^2 + sqrt(2)*(245*A*a^3 - 381*B*a^3)/a^2)*tan(1
/2*d*x + 1/2*c)^2 + 5*sqrt(2)*(73*A*a^3 - 105*B*a^3)/a^2)*tan(1/2*d*x + 1/2*c)^2 + 15*sqrt(2)*(9*A*a^3 - 17*B*
a^3)/a^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(5/2))/d