3.363 \(\int \frac{(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=192 \[ \frac{2 \left (a^3 A+9 a^2 b B+9 a A b^2+b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 (3 a B+7 A b) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (a A-b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 a A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
[Out]
(-2*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/d + (2*(a^3*A + 9*a*A*b^2 + 9*a^2*b*B +
b^3*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(7*A*b + 3*a*B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) - (2*
b^2*(a*A - b*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[
c + d*x]^(3/2))
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Rubi [A] time = 0.465498, antiderivative size = 192, normalized size of antiderivative = 1.,
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 =
{2989, 3031, 3023, 2748, 2641, 2639} \[ \frac{2 \left (a^3 A+9 a^2 b B+9 a A b^2+b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^2 (3 a B+7 A b) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (a A-b B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 a A \sin (c+d x) (a+b \cos (c+d x))^2}{3 d \cos ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]
[Out]
(-2*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/d + (2*(a^3*A + 9*a*A*b^2 + 9*a^2*b*B +
b^3*B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(7*A*b + 3*a*B)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) - (2*
b^2*(a*A - b*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d) + (2*a*A*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d*Cos[
c + d*x]^(3/2))
Rule 2989
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*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[
e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && GtQ[m, 1] && LtQ[n, -1]
Rule 3031
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]
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 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 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 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]
Rubi steps
\begin{align*} \int \frac{(a+b \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{2} a (7 A b+3 a B)+\frac{1}{2} \left (a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)-\frac{3}{2} b (a A-b B) \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (7 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{4}{3} \int \frac{-\frac{1}{4} a \left (a^2 A+10 A b^2+9 a b B\right )+\frac{3}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cos (c+d x)+\frac{3}{4} b^2 (a A-b B) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a^2 (7 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (a A-b B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{8}{9} \int \frac{-\frac{3}{8} \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right )+\frac{9}{8} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a^2 (7 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (a A-b B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \int \sqrt{\cos (c+d x)} \, dx-\frac{1}{3} \left (-a^3 A-9 a A b^2-9 a^2 b B-b^3 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 \left (a^3 A+9 a A b^2+9 a^2 b B+b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 (7 A b+3 a B) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}-\frac{2 b^2 (a A-b B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 a A (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 1.03276, size = 165, normalized size = 0.86 \[ \frac{2 \left (a^3 A+9 a^2 b B+9 a A b^2+b^3 B\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+18 a^2 A b \sin (c+d x)+2 a^3 A \tan (c+d x)+6 a^3 B \sin (c+d x)+b^3 B \sin (2 (c+d x))}{3 d \sqrt{\cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(5/2),x]
[Out]
(-6*(3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 2*(a^3*A + 9*a*A*b^
2 + 9*a^2*b*B + b^3*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 18*a^2*A*b*Sin[c + d*x] + 6*a^3*B*Sin[c
+ d*x] + b^3*B*Sin[2*(c + d*x)] + 2*a^3*A*Tan[c + d*x])/(3*d*Sqrt[Cos[c + d*x]])
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Maple [B] time = 9.1, size = 1212, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x)
[Out]
2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1
)/sin(1/2*d*x+1/2*c)^3*(8*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+2*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^3*sin(1/2*d*x+1/2*c)^2+18*A*(sin(1/2*d*x+1
/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a*b^2*sin(1/2*d*x+1/2*c)
^2+18*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^
2*b*sin(1/2*d*x+1/2*c)^2-6*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x
+1/2*c)^2-1)^(1/2)*b^3*sin(1/2*d*x+1/2*c)^2-36*A*a^2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+18*B*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^2*b*sin(1/2*d*x+1/
2*c)^2+2*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*b^3*sin(1/2*d*x+1/2*c)^2+6*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*
x+1/2*c)^2-1)^(1/2)*a^3*sin(1/2*d*x+1/2*c)^2-18*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a*b^2*sin(1/2*d*x+1/2*c)^2-12*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^4-8*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-A*a^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))-9*A*a*b^2*(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))-9*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b+3*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3+2*A*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+18*A*a^2*b*cos
(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-9*a^2*b*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)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-B*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)*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*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)*EllipticE(c
os(1/2*d*x+1/2*c),2^(1/2))*a^3+9*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)*EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2))*a*b^2+6*B*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*B*b^3*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B b^{3} \cos \left (d x + c\right )^{4} + A a^{3} +{\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
integral((B*b^3*cos(d*x + c)^4 + A*a^3 + (3*B*a*b^2 + A*b^3)*cos(d*x + c)^3 + 3*(B*a^2*b + A*a*b^2)*cos(d*x +
c)^2 + (B*a^3 + 3*A*a^2*b)*cos(d*x + c))/cos(d*x + c)^(5/2), x)
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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((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c))/cos(d*x+c)**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)