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*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)

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Rubi [A]  time = 0.446453, antiderivative size = 189, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.528046, 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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Maple [A]  time = 1.463, size = 123, normalized size = 0.7 \begin{align*}{\frac{4\,{a}^{2}\sqrt{2}}{315\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 280\,B \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+ \left ( -180\,A-900\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}+ \left ( 504\,A+1134\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+ \left ( -525\,A-735\,B \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+315\,A+315\,B \right ){\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.96898, size = 208, normalized size = 1.1 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.40363, size = 286, normalized size = 1.51 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \end{align*}

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]

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