3.820 \(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{\sqrt {b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=173 \[ \frac {2 A \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b^2 d}+\frac {6 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 b d \sqrt {\cos (c+d x)}}+\frac {2 B \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^3 d}+\frac {10 B \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 b d}+\frac {10 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}} \]

[Out]

2/5*A*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/b^2/d+2/7*B*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^3/d+10/21*B*(cos(1/2*d*x+1
/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2
)+10/21*B*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/b/d+6/5*A*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(
sin(1/2*d*x+1/2*c),2^(1/2))*(b*cos(d*x+c))^(1/2)/b/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {16, 2748, 2635, 2640, 2639, 2642, 2641} \[ \frac {2 A \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b^2 d}+\frac {6 A E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{5 b d \sqrt {\cos (c+d x)}}+\frac {2 B \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b^3 d}+\frac {10 B \sin (c+d x) \sqrt {b \cos (c+d x)}}{21 b d}+\frac {10 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*A*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*b*d*Sqrt[Cos[c + d*x]]) + (10*B*Sqrt[Cos[c + d*x]]*Ell
ipticF[(c + d*x)/2, 2])/(21*d*Sqrt[b*Cos[c + d*x]]) + (10*B*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(21*b*d) + (2*A
*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b^2*d) + (2*B*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b^3*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, 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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{\sqrt {b \cos (c+d x)}} \, dx &=\frac {\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx}{b^3}\\ &=\frac {A \int (b \cos (c+d x))^{5/2} \, dx}{b^3}+\frac {B \int (b \cos (c+d x))^{7/2} \, dx}{b^4}\\ &=\frac {2 A (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^2 d}+\frac {2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d}+\frac {(3 A) \int \sqrt {b \cos (c+d x)} \, dx}{5 b}+\frac {(5 B) \int (b \cos (c+d x))^{3/2} \, dx}{7 b^2}\\ &=\frac {10 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 A (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^2 d}+\frac {2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d}+\frac {1}{21} (5 B) \int \frac {1}{\sqrt {b \cos (c+d x)}} \, dx+\frac {\left (3 A \sqrt {b \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b \sqrt {\cos (c+d x)}}\\ &=\frac {6 A \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b d \sqrt {\cos (c+d x)}}+\frac {10 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 A (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^2 d}+\frac {2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d}+\frac {\left (5 B \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {b \cos (c+d x)}}\\ &=\frac {6 A \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b d \sqrt {\cos (c+d x)}}+\frac {10 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {10 B \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 A (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b^2 d}+\frac {2 B (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 101, normalized size = 0.58 \[ \frac {\sin (2 (c+d x)) (42 A \cos (c+d x)+15 B \cos (2 (c+d x))+65 B)+252 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+100 B \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{210 d \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(252*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 100*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (65
*B + 42*A*Cos[c + d*x] + 15*B*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(210*d*Sqrt[b*Cos[c + d*x]])

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fricas [F]  time = 1.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right )}}{b}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(d*x + c)^3 + A*cos(d*x + c)^2)*sqrt(b*cos(d*x + c))/b, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {b \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^3/sqrt(b*cos(d*x + c)), x)

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maple [A]  time = 0.96, size = 298, normalized size = 1.72 \[ -\frac {2 \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (240 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-168 A -360 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (168 A +280 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-42 A -80 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-63 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(1/2),x)

[Out]

-2/105*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-168*A-360*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(168*A+280*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)
+(-42*A-80*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-63*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^
2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+25*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-b*(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)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{\sqrt {b \cos \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^3/sqrt(b*cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)^3*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(1/2), 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+B*cos(d*x+c))/(b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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