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

Optimal. Leaf size=202 \[ \frac {2 (7 A-B) \sin (c+d x) \cos ^2(c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}-\frac {2 (7 A-31 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 a d}+\frac {4 (49 A-37 B) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 B \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}} \]

[Out]

-(A-B)*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+4/105*(49*A-37*B)*sin(
d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/35*(7*A-B)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/7*B*cos(d*x+c)
^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-2/105*(7*A-31*B)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.58, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2983, 2968, 3023, 2751, 2649, 206} \[ \frac {2 (7 A-B) \sin (c+d x) \cos ^2(c+d x)}{35 d \sqrt {a \cos (c+d x)+a}}-\frac {2 (7 A-31 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{105 a d}+\frac {4 (49 A-37 B) \sin (c+d x)}{105 d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 B \sin (c+d x) \cos ^3(c+d x)}{7 d \sqrt {a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (4*(49*A
- 37*B)*Sin[c + d*x])/(105*d*Sqrt[a + a*Cos[c + d*x]]) + (2*(7*A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(35*d*Sqrt
[a + a*Cos[c + d*x]]) + (2*B*Cos[c + d*x]^3*Sin[c + d*x])/(7*d*Sqrt[a + a*Cos[c + d*x]]) - (2*(7*A - 31*B)*Sqr
t[a + a*Cos[c + d*x]]*Sin[c + d*x])/(105*a*d)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2751

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

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2983

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*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c*(
m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (I
ntegerQ[n] || EqQ[m + 1/2, 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 \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{\sqrt {a+a \cos (c+d x)}} \, dx &=\frac {2 B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \int \frac {\cos ^2(c+d x) \left (3 a B+\frac {1}{2} a (7 A-B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{7 a}\\ &=\frac {2 (7 A-B) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {4 \int \frac {\cos (c+d x) \left (a^2 (7 A-B)-\frac {1}{4} a^2 (7 A-31 B) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{35 a^2}\\ &=\frac {2 (7 A-B) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}+\frac {4 \int \frac {a^2 (7 A-B) \cos (c+d x)-\frac {1}{4} a^2 (7 A-31 B) \cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{35 a^2}\\ &=\frac {2 (7 A-B) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (7 A-31 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 a d}+\frac {8 \int \frac {-\frac {1}{8} a^3 (7 A-31 B)+\frac {1}{4} a^3 (49 A-37 B) \cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{105 a^3}\\ &=\frac {4 (49 A-37 B) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (7 A-B) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (7 A-31 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 a d}+(-A+B) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx\\ &=\frac {4 (49 A-37 B) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (7 A-B) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (7 A-31 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 a d}+\frac {(2 (A-B)) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d}\\ &=-\frac {\sqrt {2} (A-B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {4 (49 A-37 B) \sin (c+d x)}{105 d \sqrt {a+a \cos (c+d x)}}+\frac {2 (7 A-B) \cos ^2(c+d x) \sin (c+d x)}{35 d \sqrt {a+a \cos (c+d x)}}+\frac {2 B \cos ^3(c+d x) \sin (c+d x)}{7 d \sqrt {a+a \cos (c+d x)}}-\frac {2 (7 A-31 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{105 a d}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 111, normalized size = 0.55 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (2 \sin \left (\frac {1}{2} (c+d x)\right ) ((169 B-28 A) \cos (c+d x)+6 (7 A-B) \cos (2 (c+d x))+406 A+15 B \cos (3 (c+d x))-178 B)-420 (A-B) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{210 d \sqrt {a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(-420*(A - B)*ArcTanh[Sin[(c + d*x)/2]] + 2*(406*A - 178*B + (-28*A + 169*B)*Cos[c + d*x] +
6*(7*A - B)*Cos[2*(c + d*x)] + 15*B*Cos[3*(c + d*x)])*Sin[(c + d*x)/2]))/(210*d*Sqrt[a*(1 + Cos[c + d*x])])

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fricas [A]  time = 0.61, size = 184, normalized size = 0.91 \[ \frac {4 \, {\left (15 \, B \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A - B\right )} \cos \left (d x + c\right )^{2} - {\left (7 \, A - 31 \, B\right )} \cos \left (d x + c\right ) + 91 \, A - 43 \, B\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) - \frac {105 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(4*(15*B*cos(d*x + c)^3 + 3*(7*A - B)*cos(d*x + c)^2 - (7*A - 31*B)*cos(d*x + c) + 91*A - 43*B)*sqrt(a*c
os(d*x + c) + a)*sin(d*x + c) - 105*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*log(-(cos(d*x + c)^2 - 2*sqrt
(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))
/sqrt(a))/(a*d*cos(d*x + c) + a*d)

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giac [A]  time = 1.84, size = 181, normalized size = 0.90 \[ \frac {\frac {105 \, \sqrt {2} {\left (A - B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a}} + \frac {2 \, {\left (105 \, \sqrt {2} A a^{3} + {\left ({\left (\sqrt {2} {\left (119 \, A a^{3} - 92 \, B a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7 \, \sqrt {2} {\left (37 \, A a^{3} - 16 \, B a^{3}\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, \sqrt {2} {\left (7 \, A a^{3} - 4 \, B a^{3}\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}}{105 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(105*sqrt(2)*(A - B)*log(abs(-sqrt(a)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/sqrt(a
) + 2*(105*sqrt(2)*A*a^3 + ((sqrt(2)*(119*A*a^3 - 92*B*a^3)*tan(1/2*d*x + 1/2*c)^2 + 7*sqrt(2)*(37*A*a^3 - 16*
B*a^3))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(7*A*a^3 - 4*B*a^3))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/
(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2))/d

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maple [A]  time = 0.83, size = 281, normalized size = 1.39 \[ \frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-240 B \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (A +2 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-140 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (A +2 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-105 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a A +105 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a B +210 A \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{105 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-240*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)
*sin(1/2*d*x+1/2*c)^6+168*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*(A+2*B)*sin(1/2*d*x+1/2*c)^4-140*2^(1
/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*(A+2*B)*sin(1/2*d*x+1/2*c)^2-105*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a))*a*A+105*2^(1/2)*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1/2*
c)^2)^(1/2)+a))*a*B+210*A*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(3/2)/sin(1/2*d*x+1/2*c)/(a*cos(1/
2*d*x+1/2*c)^2)^(1/2)/d

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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