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

Optimal. Leaf size=145 \[ \frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {7 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{6 a^2 d}-\frac {\sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {13 \sin (c+d x)}{3 a d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-1/2*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)+11/4*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x
+c))^(1/2))/a^(3/2)/d*2^(1/2)-13/3*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+7/6*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)
/a^2/d

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Rubi [A]  time = 0.26, antiderivative size = 145, normalized size of antiderivative = 1.00, 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, 3023, 2751, 2649, 206} \[ \frac {7 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{6 a^2 d}+\frac {11 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\sin (c+d x) \cos ^2(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {13 \sin (c+d x)}{3 a d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - (Cos[c + d*x]^
2*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) - (13*Sin[c + d*x])/(3*a*d*Sqrt[a + a*Cos[c + d*x]]) + (7*Sqr
t[a + a*Cos[c + d*x]]*Sin[c + d*x])/(6*a^2*d)

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])

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 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 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 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]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {\cos (c+d x) \left (2 a-\frac {7}{2} a \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {\int \frac {2 a \cos (c+d x)-\frac {7}{2} a \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {7 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}-\frac {\int \frac {-\frac {7 a^2}{4}+\frac {13}{2} a^2 \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{3 a^3}\\ &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {13 \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}+\frac {7 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}+\frac {11 \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}\\ &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {13 \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}+\frac {7 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}-\frac {11 \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 \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 {\cos ^2(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}-\frac {13 \sin (c+d x)}{3 a d \sqrt {a+a \cos (c+d x)}}+\frac {7 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{6 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.96, size = 196, normalized size = 1.35 \[ \frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (-72 \sin \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )+8 \sin \left (\frac {3 c}{2}\right ) \cos \left (\frac {3 d x}{2}\right )-72 \cos \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+8 \cos \left (\frac {3 c}{2}\right ) \sin \left (\frac {3 d x}{2}\right )-\frac {3}{\left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )^2}+\frac {3}{\left (\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )\right )^2}-66 \log \left (\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )\right )+66 \log \left (\sin \left (\frac {1}{4} (c+d x)\right )+\cos \left (\frac {1}{4} (c+d x)\right )\right )\right )}{6 d (a (\cos (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*(-66*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] + 66*Log[Cos[(c + d*x)/4] + Sin[(c + d*x)/4]
] - 72*Cos[(d*x)/2]*Sin[c/2] + 8*Cos[(3*d*x)/2]*Sin[(3*c)/2] - 72*Cos[c/2]*Sin[(d*x)/2] + 8*Cos[(3*c)/2]*Sin[(
3*d*x)/2] - 3/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4])^2 + 3/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4])^2))/(6*d*(a*(1
+ Cos[c + d*x]))^(3/2))

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fricas [A]  time = 0.68, size = 174, normalized size = 1.20 \[ \frac {33 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (4 \, \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right ) - 19\right )} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(33*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*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*sqrt(a*
cos(d*x + c) + a)*(4*cos(d*x + c)^2 - 12*cos(d*x + c) - 19)*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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giac [A]  time = 1.35, size = 115, normalized size = 0.79 \[ -\frac {\frac {{\left ({\left (3 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 46 \, \sqrt {2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 27 \, \sqrt {2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} + \frac {33 \, \sqrt {2} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(((3*sqrt(2)*tan(1/2*d*x + 1/2*c)^2 + 46*sqrt(2))*tan(1/2*d*x + 1/2*c)^2 + 27*sqrt(2))*tan(1/2*d*x + 1/2
*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) + 33*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^(3/2))/d

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maple [A]  time = 0.32, size = 234, normalized size = 1.61 \[ \frac {\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (16 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-33 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-27 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+33 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(
1/2*d*x+1/2*c)^4+8*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2-33*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)^2-27*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+33*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^(5/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x
+1/2*c)^2)^(1/2)/d

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^3}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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