3.83 \(\int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=189 \[ \frac {2 a^2 (9 A+10 B) \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (39 A+34 B) \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (39 A+34 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}-\frac {4 a (39 A+34 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d} \]

[Out]

2/105*(39*A+34*B)*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/45*a^2*(39*A+34*B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
+2/63*a^2*(9*A+10*B)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-4/315*a*(39*A+34*B)*sin(d*x+c)*(a+a*cos(
d*x+c))^(1/2)/d+2/9*a*B*cos(d*x+c)^3*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.45, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2976, 2981, 2759, 2751, 2646} \[ \frac {2 a^2 (9 A+10 B) \sin (c+d x) \cos ^3(c+d x)}{63 d \sqrt {a \cos (c+d x)+a}}+\frac {2 a^2 (39 A+34 B) \sin (c+d x)}{45 d \sqrt {a \cos (c+d x)+a}}+\frac {2 (39 A+34 B) \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{105 d}-\frac {4 a (39 A+34 B) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{315 d}+\frac {2 a B \sin (c+d x) \cos ^3(c+d x) \sqrt {a \cos (c+d x)+a}}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]),x]

[Out]

(2*a^2*(39*A + 34*B)*Sin[c + d*x])/(45*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(9*A + 10*B)*Cos[c + d*x]^3*Sin[c
+ d*x])/(63*d*Sqrt[a + a*Cos[c + d*x]]) - (4*a*(39*A + 34*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(315*d) +
(2*a*B*Cos[c + d*x]^3*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(9*d) + (2*(39*A + 34*B)*(a + a*Cos[c + d*x])^(3/
2)*Sin[c + d*x])/(105*d)

Rule 2646

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(-2*b*Cos[c + d*x])/(d*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 2759

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

Rule 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx &=\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2}{9} \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \left (\frac {3}{2} a (3 A+2 B)+\frac {1}{2} a (9 A+10 B) \cos (c+d x)\right ) \, dx\\ &=\frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {1}{21} (a (39 A+34 B)) \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {1}{105} (2 (39 A+34 B)) \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}+\frac {1}{45} (a (39 A+34 B)) \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {2 a^2 (39 A+34 B) \sin (c+d x)}{45 d \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (9 A+10 B) \cos ^3(c+d x) \sin (c+d x)}{63 d \sqrt {a+a \cos (c+d x)}}-\frac {4 a (39 A+34 B) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{315 d}+\frac {2 a B \cos ^3(c+d x) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d}+\frac {2 (39 A+34 B) (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{105 d}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 103, normalized size = 0.54 \[ \frac {a \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\cos (c+d x)+1)} (2 (759 A+799 B) \cos (c+d x)+(468 A+548 B) \cos (2 (c+d x))+90 A \cos (3 (c+d x))+2964 A+170 B \cos (3 (c+d x))+35 B \cos (4 (c+d x))+2689 B)}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + a*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x]),x]

[Out]

(a*Sqrt[a*(1 + Cos[c + d*x])]*(2964*A + 2689*B + 2*(759*A + 799*B)*Cos[c + d*x] + (468*A + 548*B)*Cos[2*(c + d
*x)] + 90*A*Cos[3*(c + d*x)] + 170*B*Cos[3*(c + d*x)] + 35*B*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/(1260*d)

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fricas [A]  time = 0.59, size = 107, normalized size = 0.57 \[ \frac {2 \, {\left (35 \, B a \cos \left (d x + c\right )^{4} + 5 \, {\left (9 \, A + 17 \, B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right )^{2} + 4 \, {\left (39 \, A + 34 \, B\right )} a \cos \left (d x + c\right ) + 8 \, {\left (39 \, A + 34 \, B\right )} a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*B*a*cos(d*x + c)^4 + 5*(9*A + 17*B)*a*cos(d*x + c)^3 + 3*(39*A + 34*B)*a*cos(d*x + c)^2 + 4*(39*A +
34*B)*a*cos(d*x + c) + 8*(39*A + 34*B)*a)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/(d*cos(d*x + c) + d)

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giac [A]  time = 1.84, size = 245, normalized size = 1.30 \[ \frac {1}{2520} \, \sqrt {2} {\left (\frac {35 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )}{d} + \frac {45 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )}{d} + \frac {378 \, {\left (A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {1050 \, {\left (A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {630 \, {\left (3 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 4 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d} + \frac {1260 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*sqrt(2)*(35*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(9/2*d*x + 9/2*c)/d + 45*(2*A*a*sgn(cos(1/2*d*x + 1/2*c))
+ 3*B*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(7/2*d*x + 7/2*c)/d + 378*(A*a*sgn(cos(1/2*d*x + 1/2*c)) + B*a*sgn(cos(1
/2*d*x + 1/2*c)))*sin(5/2*d*x + 5/2*c)/d + 1050*(A*a*sgn(cos(1/2*d*x + 1/2*c)) + B*a*sgn(cos(1/2*d*x + 1/2*c))
)*sin(3/2*d*x + 3/2*c)/d + 630*(3*A*a*sgn(cos(1/2*d*x + 1/2*c)) + 4*B*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x
+ 1/2*c)/d + 1260*(2*A*a*sgn(cos(1/2*d*x + 1/2*c)) + B*a*sgn(cos(1/2*d*x + 1/2*c)))*sin(1/2*d*x + 1/2*c)/d)*s
qrt(a)

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maple [A]  time = 0.42, size = 123, normalized size = 0.65 \[ \frac {4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (280 B \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-180 A -900 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (504 A +1134 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-525 A -735 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+315 A +315 B \right ) \sqrt {2}}{315 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*cos(1/2*d*x+1/2*c)*a^2*sin(1/2*d*x+1/2*c)*(280*B*sin(1/2*d*x+1/2*c)^8+(-180*A-900*B)*sin(1/2*d*x+1/2*c)^
6+(504*A+1134*B)*sin(1/2*d*x+1/2*c)^4+(-525*A-735*B)*sin(1/2*d*x+1/2*c)^2+315*A+315*B)*2^(1/2)/(a*cos(1/2*d*x+
1/2*c)^2)^(1/2)/d

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maxima [A]  time = 0.79, size = 154, normalized size = 0.81 \[ \frac {6 \, {\left (15 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 63 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 175 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 735 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A \sqrt {a} + {\left (35 \, \sqrt {2} a \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 135 \, \sqrt {2} a \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 378 \, \sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 1050 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3780 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*(6*(15*sqrt(2)*a*sin(7/2*d*x + 7/2*c) + 63*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 175*sqrt(2)*a*sin(3/2*d*x +
3/2*c) + 735*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + (35*sqrt(2)*a*sin(9/2*d*x + 9/2*c) + 135*sqrt(2)*a*s
in(7/2*d*x + 7/2*c) + 378*sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 1050*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 3780*sqrt(2)*
a*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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