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

Optimal. Leaf size=159 \[ \frac {2 b \left (21 a^2+5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a \left (5 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {32 a b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d}+\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}{7 d} \]

[Out]

2/5*a*(5*a^2+9*b^2)*(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))/d+2/
21*b*(21*a^2+5*b^2)*(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))/d+32
/35*a*b^2*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*b^2*cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))*sin(d*x+c)/d+2/21*b*(21*a^2+
5*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.20, antiderivative size = 159, 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 = {2793, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 b \left (21 a^2+5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a \left (5 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {32 a b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 d}+\frac {2 b^2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(5*a^2 + 9*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*b*(21*a^2 + 5*b^2)*EllipticF[(c + d*x)/2, 2])/(21*d
) + (2*b*(21*a^2 + 5*b^2)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (32*a*b^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])
/(35*d) + (2*b^2*Cos[c + d*x]^(3/2)*(a + b*Cos[c + d*x])*Sin[c + d*x])/(7*d)

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 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 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 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 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 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \, dx &=\frac {2 b^2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sqrt {\cos (c+d x)} \left (\frac {1}{2} a \left (7 a^2+3 b^2\right )+\frac {1}{2} b \left (21 a^2+5 b^2\right ) \cos (c+d x)+8 a b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac {32 a b^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\cos (c+d x)} \left (\frac {7}{4} a \left (5 a^2+9 b^2\right )+\frac {5}{4} b \left (21 a^2+5 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {32 a b^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{7} \left (b \left (21 a^2+5 b^2\right )\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{5} \left (a \left (5 a^2+9 b^2\right )\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 a \left (5 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {32 a b^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}+\frac {1}{21} \left (b \left (21 a^2+5 b^2\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a \left (5 a^2+9 b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 b \left (21 a^2+5 b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 b \left (21 a^2+5 b^2\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {32 a b^2 \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 d}+\frac {2 b^2 \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]  time = 0.79, size = 110, normalized size = 0.69 \[ \frac {42 \left (5 a^3+9 a b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 \left (21 a^2 b+5 b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+b \sin (c+d x) \sqrt {\cos (c+d x)} \left (210 a^2+126 a b \cos (c+d x)+15 b^2 \cos (2 (c+d x))+65 b^2\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(42*(5*a^3 + 9*a*b^2)*EllipticE[(c + d*x)/2, 2] + 10*(21*a^2*b + 5*b^3)*EllipticF[(c + d*x)/2, 2] + b*Sqrt[Cos
[c + d*x]]*(210*a^2 + 65*b^2 + 126*a*b*Cos[c + d*x] + 15*b^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(105*d)

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fricas [F]  time = 1.15, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3)*sqrt(cos(d*x + c)), x)

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

[Out]

Timed out

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maple [B]  time = 0.74, size = 421, normalized size = 2.65 \[ -\frac {2 \sqrt {\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^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-504 b^{2} a -360 b^{3}\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 (420 a^{2} b +504 b^{2} a +280 b^{3}\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 (-210 a^{2} b -126 b^{2} a -80 b^{3}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+105 a^{2} 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 )+25 b^{3} \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 )-105 \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 ) a^{3}-189 \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 ) a \,b^{2}\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8+(-504*a*b^2-360*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(420*a^2*b+504*a*b^2+280*b^3)*sin(1/2*d*x+1/2*c
)^4*cos(1/2*d*x+1/2*c)+(-210*a^2*b-126*a*b^2-80*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+105*a^2*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))+25*b^3*(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))-105*(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))*a^3-189*(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))*a*b^2)/(-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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cos(d*x + c) + a)^3*sqrt(cos(d*x + c)), x)

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mupad [B]  time = 0.94, size = 146, normalized size = 0.92 \[ \frac {2\,\left (a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+a^2\,b\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}-\frac {2\,b^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,a\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a^3*ellipticE(c/2 + (d*x)/2, 2) + a^2*b*ellipticF(c/2 + (d*x)/2, 2) + a^2*b*cos(c + d*x)^(1/2)*sin(c + d*x
)))/d - (2*b^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)
^2)^(1/2)) - (6*a*b^2*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c
+ d*x)^2)^(1/2))

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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)**(1/2)*(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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