3.161 \(\int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\)
Optimal. Leaf size=180 \[ \frac{(A+3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A+9 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(A-6 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2} \]
[Out]
-((A + 9*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((A + 3*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) + ((A - B)
*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((A - 6*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
15*a*d*(a + a*Cos[c + d*x])^2) + ((A + 9*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[c + d*x]))
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Rubi [A] time = 0.470111, antiderivative size = 180, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used =
{2977, 2978, 2748, 2641, 2639} \[ \frac{(A+3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}-\frac{(A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A+9 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{(A-B) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(A-6 B) \sin (c+d x) \sqrt{\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]
[Out]
-((A + 9*B)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((A + 3*B)*EllipticF[(c + d*x)/2, 2])/(6*a^3*d) + ((A - B)
*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((A - 6*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(
15*a*d*(a + a*Cos[c + d*x])^2) + ((A + 9*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[c + d*x]))
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 2978
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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[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] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])
Rule 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rubi steps
\begin{align*} \int \frac{\cos ^{\frac{3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3}{2} a (A-B)+\frac{1}{2} a (A+9 B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A-6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a^2 (A-6 B)+\frac{1}{2} a^2 (4 A+21 B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A-6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(A+9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \frac{\frac{5}{4} a^3 (A+3 B)-\frac{3}{4} a^3 (A+9 B) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a^6}\\ &=\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A-6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(A+9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{(A+3 B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}-\frac{(A+9 B) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}\\ &=-\frac{(A+9 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{(A+3 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(A-B) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(A-6 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{(A+9 B) \sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 6.69926, size = 1265, normalized size = 7.03 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]
[Out]
((-I/10)*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I
)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*
x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1
+ E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[
(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c]
+ I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/(a + a*Cos[
c + d*x])^3 - (((9*I)/10)*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/
4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x)
)*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))
*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*
Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*
I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[
c])))/(a + a*Cos[c + d*x])^3 - (2*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x
- ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + C
ot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^3*S
qrt[1 + Cot[c]^2]) - (2*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[
Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*S
in[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d*x])^3*Sqrt[1 + Cot[
c]^2]) + (Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*((4*(A + 9*B)*Csc[c])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]
^3*(4*A*Sin[(d*x)/2] - 9*B*Sin[(d*x)/2]))/(15*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d
*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 9*B*Sin[(d*x)/2]))/(5*d) + (4*(4*A - 9*B)*Se
c[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) - (2*(A - B)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(a + a*Cos[c + d*x])^3
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Maple [B] time = 3.637, size = 451, normalized size = 2.5 \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)^(3/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3,x)
[Out]
-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2*d*x+1/2*c)^8+10*A*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+6*A*
cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*
c),2^(1/2))+108*B*cos(1/2*d*x+1/2*c)^8+30*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ell
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^5+54*B*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)
*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*A*cos(1/2*d*x+1/2*c)^6-198*B*cos(1/
2*d*x+1/2*c)^6-24*A*cos(1/2*d*x+1/2*c)^4+114*B*cos(1/2*d*x+1/2*c)^4+17*A*cos(1/2*d*x+1/2*c)^2-27*B*cos(1/2*d*x
+1/2*c)^2-3*A+3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1
/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^3, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
[Out]
integral((B*cos(d*x + c)^2 + A*cos(d*x + c))*sqrt(cos(d*x + c))/(a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3
*a^3*cos(d*x + c) + a^3), x)
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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)**(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*cos(d*x + c) + a)^3, x)