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

Optimal. Leaf size=233 \[ \frac{a^2 (48 A+56 B+39 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{3/2} (112 A+88 B+75 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^2 (112 A+88 B+75 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{a (8 B+3 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{24 d}+\frac{C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d} \]

[Out]

(a^(3/2)*(112*A + 88*B + 75*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^2*(112*A +
88*B + 75*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(48*A + 56*B + 39*C)*Cos
[c + d*x]^(3/2)*Sin[c + d*x])/(96*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(8*B + 3*C)*Cos[c + d*x]^(3/2)*Sqrt[a + a*C
os[c + d*x]]*Sin[c + d*x])/(24*d) + (C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.64485, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3045, 2976, 2981, 2770, 2774, 216} \[ \frac{a^2 (48 A+56 B+39 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{96 d \sqrt{a \cos (c+d x)+a}}+\frac{a^{3/2} (112 A+88 B+75 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{64 d}+\frac{a^2 (112 A+88 B+75 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{64 d \sqrt{a \cos (c+d x)+a}}+\frac{a (8 B+3 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \cos (c+d x)+a}}{24 d}+\frac{C \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(3/2)*(112*A + 88*B + 75*C)*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(64*d) + (a^2*(112*A +
88*B + 75*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(64*d*Sqrt[a + a*Cos[c + d*x]]) + (a^2*(48*A + 56*B + 39*C)*Cos
[c + d*x]^(3/2)*Sin[c + d*x])/(96*d*Sqrt[a + a*Cos[c + d*x]]) + (a*(8*B + 3*C)*Cos[c + d*x]^(3/2)*Sqrt[a + a*C
os[c + d*x]]*Sin[c + d*x])/(24*d) + (C*Cos[c + d*x]^(3/2)*(a + a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(4*d)

Rule 3045

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

Rule 2976

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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, 0] && IntegerQ[2*n]

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 \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{\int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^{3/2} \left (\frac{1}{2} a (8 A+3 C)+\frac{1}{2} a (8 B+3 C) \cos (c+d x)\right ) \, dx}{4 a}\\ &=\frac{a (8 B+3 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{24 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{\int \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)} \left (\frac{3}{4} a^2 (16 A+8 B+9 C)+\frac{1}{4} a^2 (48 A+56 B+39 C) \cos (c+d x)\right ) \, dx}{12 a}\\ &=\frac{a^2 (48 A+56 B+39 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a (8 B+3 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{24 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{1}{64} (a (112 A+88 B+75 C)) \int \sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)} \, dx\\ &=\frac{a^2 (112 A+88 B+75 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (48 A+56 B+39 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a (8 B+3 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{24 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac{1}{128} (a (112 A+88 B+75 C)) \int \frac{\sqrt{a+a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{a^2 (112 A+88 B+75 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (48 A+56 B+39 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a (8 B+3 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{24 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d}-\frac{(a (112 A+88 B+75 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 )}{64 d}\\ &=\frac{a^{3/2} (112 A+88 B+75 C) \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{64 d}+\frac{a^2 (112 A+88 B+75 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{64 d \sqrt{a+a \cos (c+d x)}}+\frac{a^2 (48 A+56 B+39 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{96 d \sqrt{a+a \cos (c+d x)}}+\frac{a (8 B+3 C) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{24 d}+\frac{C \cos ^{\frac{3}{2}}(c+d x) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.950164, size = 145, normalized size = 0.62 \[ \frac{a \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} \left (3 \sqrt{2} (112 A+88 B+75 C) \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} (2 (48 A+88 B+93 C) \cos (c+d x)+336 A+4 (8 B+15 C) \cos (2 (c+d x))+296 B+12 C \cos (3 (c+d x))+285 C)\right )}{384 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(3*Sqrt[2]*(112*A + 88*B + 75*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2
]] + 2*Sqrt[Cos[c + d*x]]*(336*A + 296*B + 285*C + 2*(48*A + 88*B + 93*C)*Cos[c + d*x] + 4*(8*B + 15*C)*Cos[2*
(c + d*x)] + 12*C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(384*d)

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Maple [B]  time = 0.112, size = 623, normalized size = 2.7 \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)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

-1/192/d*a*(-1+cos(d*x+c))^3*(96*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+528*A*sin(d*x+c)*
cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+64*B*cos(d*x+c)^4*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+
768*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+240*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(1+cos
(d*x+c)))^(3/2)+48*C*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+336*A*sin(d*x+c)*(cos(d*x+c)/(1
+cos(d*x+c)))^(5/2)+440*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+120*C*sin(d*x+c)*cos(d*x+c
)^4*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+264*B*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+150*C*sin(
d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+225*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)+336*A*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))+264*B*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))+225*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)))*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(5
/2)/sin(d*x+c)^6

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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)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 6.81902, size = 493, normalized size = 2.12 \begin{align*} \frac{{\left (48 \, C a \cos \left (d x + c\right )^{3} + 8 \,{\left (8 \, B + 15 \, C\right )} a \cos \left (d x + c\right )^{2} + 2 \,{\left (48 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right ) + 3 \,{\left (112 \, A + 88 \, B + 75 \, C\right )} a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 3 \,{\left ({\left (112 \, A + 88 \, B + 75 \, C\right )} a \cos \left (d x + c\right ) +{\left (112 \, A + 88 \, B + 75 \, C\right )} a\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 )}{192 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/192*((48*C*a*cos(d*x + c)^3 + 8*(8*B + 15*C)*a*cos(d*x + c)^2 + 2*(48*A + 88*B + 75*C)*a*cos(d*x + c) + 3*(1
12*A + 88*B + 75*C)*a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 3*((112*A + 88*B + 75*C)*a*c
os(d*x + c) + (112*A + 88*B + 75*C)*a)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin
(d*x + c))))/(d*cos(d*x + c) + 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)**(1/2)*(a+a*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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Giac [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)^(1/2)*(a+a*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

Timed out