3.128 \(\int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) (A+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=134 \[ \frac {2 a (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (7 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {2 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d} \]

[Out]

2/5*a*(5*A+3*C)*(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*a
*(7*A+5*C)*(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+2/5*a*C*cos
(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*a*C*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/21*a*(7*A+5*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/
d

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Rubi [A]  time = 0.19, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3034, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 a (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (7 A+5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 a C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}+\frac {2 a C \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(7*A + 5*C)*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a*
(7*A + 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*a*C*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*C*C
os[c + d*x]^(5/2)*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 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 3034

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m
+ 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*d*(C*(m + 2) + A*(m
+ 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {2 a C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {2}{7} \int \sqrt {\cos (c+d x)} \left (\frac {7 a A}{2}+\frac {1}{2} a (7 A+5 C) \cos (c+d x)+\frac {7}{2} a C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {4}{35} \int \sqrt {\cos (c+d x)} \left (\frac {7}{4} a (5 A+3 C)+\frac {5}{4} a (7 A+5 C) \cos (c+d x)\right ) \, dx\\ &=\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{5} (a (5 A+3 C)) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{7} (a (7 A+5 C)) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (7 A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac {1}{21} (a (7 A+5 C)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a (7 A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a C \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [C]  time = 6.26, size = 872, normalized size = 6.51 \[ a \left (\sqrt {\cos (c+d x)} (\cos (c+d x)+1) \left (-\frac {(5 A+3 C) \cot (c)}{5 d}+\frac {(28 A+23 C) \cos (d x) \sin (c)}{84 d}+\frac {C \cos (2 d x) \sin (2 c)}{10 d}+\frac {C \cos (3 d x) \sin (3 c)}{28 d}+\frac {(28 A+23 C) \cos (c) \sin (d x)}{84 d}+\frac {C \cos (2 c) \sin (2 d x)}{10 d}+\frac {C \cos (3 c) \sin (3 d x)}{28 d}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )-\frac {A (\cos (c+d x)+1) \csc (c) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt {\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}} \sqrt {\tan ^2(c)+1}}-\frac {\frac {2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac {\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {\tan ^2(c)+1}}}{\sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}-\frac {3 C (\cos (c+d x)+1) \csc (c) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (d x+\tan ^{-1}(\tan (c))\right )\right ) \sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {1-\cos \left (d x+\tan ^{-1}(\tan (c))\right )} \sqrt {\cos \left (d x+\tan ^{-1}(\tan (c))\right )+1} \sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}} \sqrt {\tan ^2(c)+1}}-\frac {\frac {2 \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1} \cos ^2(c)}{\cos ^2(c)+\sin ^2(c)}+\frac {\sin \left (d x+\tan ^{-1}(\tan (c))\right ) \tan (c)}{\sqrt {\tan ^2(c)+1}}}{\sqrt {\cos (c) \cos \left (d x+\tan ^{-1}(\tan (c))\right ) \sqrt {\tan ^2(c)+1}}}\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{10 d}-\frac {A (\cos (c+d x)+1) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {-\sqrt {\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{3 d \sqrt {\cot ^2(c)+1}}-\frac {5 C (\cos (c+d x)+1) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt {1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {-\sqrt {\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt {\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{21 d \sqrt {\cot ^2(c)+1}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(-1/5*((5*A + 3*C)*Cot[c])/d + ((28*A + 23*C)*Co
s[d*x]*Sin[c])/(84*d) + (C*Cos[2*d*x]*Sin[2*c])/(10*d) + (C*Cos[3*d*x]*Sin[3*c])/(28*d) + ((28*A + 23*C)*Cos[c
]*Sin[d*x])/(84*d) + (C*Cos[2*c]*Sin[2*d*x])/(10*d) + (C*Cos[3*c]*Sin[3*d*x])/(28*d)) - (A*(1 + Cos[c + d*x])*
Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan
[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqr
t[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (5*C*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[
{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*
x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Co
t[c]]]])/(21*d*Sqrt[1 + Cot[c]^2]) - (A*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1
/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Ta
n[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 +
Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*S
qrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d) -
(3*C*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[
Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[T
an[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Ta
n[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[
c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(10*d))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C a \cos \left (d x + c\right )^{3} + C a \cos \left (d x + c\right )^{2} + A a \cos \left (d x + c\right ) + A a\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)*sqrt(cos(d*x + c)), x)

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maple [B]  time = 1.47, size = 378, normalized size = 2.82 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a \left (240 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-528 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (140 A +448 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-70 A -122 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+35 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))*(A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),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)*a*(240*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8-528*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(140*A+448*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A-1
22*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+35*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*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)*E
llipticE(cos(1/2*d*x+1/2*c),2^(1/2))+25*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-63*C*(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(c
os(1/2*d*x+1/2*c),2^(1/2)))/(-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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*(a*cos(d*x + c) + a)*sqrt(cos(d*x + c)), x)

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mupad [B]  time = 0.39, size = 139, normalized size = 1.04 \[ \frac {2\,A\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,C\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(1/2)*(A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x)),x)

[Out]

(2*A*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3*d) + (2*A*a*ellipticE(c/2 + (d*x)/2
, 2))/d - (2*C*a*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)) - (2*C*a*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))

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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((a+a*cos(d*x+c))*(A+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2),x)

[Out]

Timed out

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