3.123 \(\int \frac{\cos ^3(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=140 \[ \frac{2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}-\frac{2 \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 a d}+\frac{28 \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (28*Sin[c + d*x]
)/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) - (2*Sqrt[a
+ a*Cos[c + d*x]]*Sin[c + d*x])/(15*a*d)

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Rubi [A]  time = 0.239252, antiderivative size = 140, 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 = {2778, 2968, 3023, 2751, 2649, 206} \[ \frac{2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt{a \cos (c+d x)+a}}-\frac{2 \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 a d}+\frac{28 \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (28*Sin[c + d*x]
)/(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) - (2*Sqrt[a
+ a*Cos[c + d*x]]*Sin[c + d*x])/(15*a*d)

Rule 2778

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-2*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/(b*(2*n
- 1)), Int[((c + d*Sin[e + f*x])^(n - 2)*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n -
3))*Sin[e + f*x], x])/Sqrt[a + b*Sin[e + f*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] && GtQ[n, 1] && IntegerQ[2*n]

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)}{\sqrt{a+a \cos (c+d x)}} \, dx &=\frac{2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{\int \frac{\cos (c+d x) (-4 a+a \cos (c+d x))}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac{2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{\int \frac{-4 a \cos (c+d x)+a \cos ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{5 a}\\ &=\frac{2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}-\frac{2 \int \frac{\frac{a^2}{2}-7 a^2 \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{15 a^2}\\ &=\frac{28 \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}-\int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx\\ &=\frac{28 \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}+\frac{2 \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 )}{d}\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{\sqrt{a} d}+\frac{28 \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt{a+a \cos (c+d x)}}-\frac{2 \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 a d}\\ \end{align*}

Mathematica [A]  time = 0.164344, size = 118, normalized size = 0.84 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (60 \sin \left (\frac{1}{2} (c+d x)\right )-5 \sin \left (\frac{3}{2} (c+d x)\right )+3 \sin \left (\frac{5}{2} (c+d x)\right )+30 \log \left (\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right )-30 \log \left (\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )\right )\right )}{15 d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(Cos[(c + d*x)/2]*(30*Log[Cos[(c + d*x)/4] - Sin[(c + d*x)/4]] - 30*Log[Cos[(c + d*x)/4] + Sin[(c + d*x)/4]] +
60*Sin[(c + d*x)/2] - 5*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c + d*x))/2]))/(15*d*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [A]  time = 1.327, size = 183, normalized size = 1.3 \begin{align*}{\frac{\sqrt{2}}{15\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 24\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-20\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+30\,\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-15\,\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 ){a}^{-{\frac{3}{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)^(1/2),x)

[Out]

1/15*cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(
1/2*d*x+1/2*c)^4-20*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+30*a^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)-15*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^(3/2)/sin(1/2*d*x+1/2*
c)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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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))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.63261, size = 397, normalized size = 2.84 \begin{align*} \frac{4 \, \sqrt{a \cos \left (d x + c\right ) + a}{\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\right )} \sin \left (d x + c\right ) + \frac{15 \, \sqrt{2}{\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} + \frac{2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}}}{30 \,{\left (a d \cos \left (d x + c\right ) + a 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))^(1/2),x, algorithm="fricas")

[Out]

1/30*(4*sqrt(a*cos(d*x + c) + a)*(3*cos(d*x + c)^2 - cos(d*x + c) + 13)*sin(d*x + c) + 15*sqrt(2)*(a*cos(d*x +
c) + a)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/
(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*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/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 2.62716, size = 157, normalized size = 1.12 \begin{align*} \frac{\sqrt{2}{\left (\frac{15 \, \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 )}{\sqrt{a}} + \frac{2 \,{\left ({\left (17 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 \, a^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 \, 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}}}\right )}}{15 \, 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))^(1/2),x, algorithm="giac")

[Out]

1/15*sqrt(2)*(15*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/sqrt(a) + 2*((17
*a^2*tan(1/2*d*x + 1/2*c)^2 + 20*a^2)*tan(1/2*d*x + 1/2*c)^2 + 15*a^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1
/2*c)^2 + a)^(5/2))/d