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

Optimal. Leaf size=145 \[ \frac{9 \sin (c+d x)}{4 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{75 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{13 \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

[Out]

(-75*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Cos[c + d*x
]^2*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + (13*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + (
9*Sin[c + d*x])/(4*a^2*d*Sqrt[a + a*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.270186, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2765, 2968, 3019, 2751, 2649, 206} \[ \frac{9 \sin (c+d x)}{4 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{75 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}+\frac{13 \sin (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-75*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (Cos[c + d*x
]^2*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + (13*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)) + (
9*Sin[c + d*x])/(4*a^2*d*Sqrt[a + a*Cos[c + d*x]])

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(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 - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], 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] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 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 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

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+a \cos (c+d x))^{5/2}} \, dx &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{\int \frac{\cos (c+d x) \left (2 a-\frac{9}{2} a \cos (c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{\int \frac{2 a \cos (c+d x)-\frac{9}{2} a \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{-\frac{39 a^2}{4}+9 a^2 \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{9 \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{75 \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{9 \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{75 \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 )}{16 a^2 d}\\ &=-\frac{75 \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{\cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{9 \sin (c+d x)}{4 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 4.38222, size = 216, normalized size = 1.49 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (128 \sin \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )+128 \cos \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+\frac{21}{\left (\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right )^2}-\frac{21}{\left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )^2}-\frac{1}{\left (\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right )^4}+\frac{1}{\left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )^4}+150 \log \left (\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right )-150 \log \left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )\right )}{8 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(150*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] - 150*Log[Cos[(c + d*x)/4] + Sin[(c + d*x)/4
]] + 128*Cos[(d*x)/2]*Sin[c/2] + 128*Cos[c/2]*Sin[(d*x)/2] - (Cos[(c + d*x)/4] - Sin[(c + d*x)/4])^(-4) + 21/(
Cos[(c + d*x)/4] - Sin[(c + d*x)/4])^2 + (Cos[(c + d*x)/4] + Sin[(c + d*x)/4])^(-4) - 21/(Cos[(c + d*x)/4] + S
in[(c + d*x)/4])^2))/(8*d*(a*(1 + Cos[c + d*x]))^(5/2))

________________________________________________________________________________________

Maple [A]  time = 1.507, size = 208, normalized size = 1.4 \begin{align*} -{\frac{1}{32\,d}\sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 75\,\sqrt{2}\ln \left ( 2\,{\frac{2\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+2\,a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-64\,\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-21\,\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{a}^{-{\frac{7}{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+cos(d*x+c)*a)^(5/2),x)

[Out]

-1/32*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(75*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d
*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-64*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-21*2^
(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^2+2*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(
1/2))/cos(1/2*d*x+1/2*c)^3/a^(7/2)/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+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.68844, size = 532, normalized size = 3.67 \begin{align*} \frac{75 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\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 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (32 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 49\right )} \sin \left (d x + c\right )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(75*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*s
qrt(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*sqrt(a*cos(d*x + c) + a)*(32*cos(d*x + c)^2 + 85*cos(d*x + c) + 49)*sin(d*x + c))/(a^3*d*cos(d*x +
c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*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+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.7545, size = 167, normalized size = 1.15 \begin{align*} -\frac{\frac{{\left ({\left (\frac{2 \, \sqrt{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{2}} - \frac{17 \, \sqrt{2}}{a^{2}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{83 \, \sqrt{2}}{a^{2}}\right )} \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}} - \frac{75 \, \sqrt{2} \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{5}{2}}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(((2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/a^2 - 17*sqrt(2)/a^2)*tan(1/2*d*x + 1/2*c)^2 - 83*sqrt(2)/a^2)*tan(1
/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) - 75*sqrt(2)*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^(5/2))/d