3.871 \(\int \frac{(a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=202 \[ \frac{2 \left (9 a^2 b B+3 a^3 C+3 a b^2 C+b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (-15 a^2 b C+5 a^3 B-15 a b^2 B-3 b^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}-\frac{2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}} \]
[Out]
(-2*(5*a^3*B - 15*a*b^2*B - 15*a^2*b*C - 3*b^3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(9*a^2*b*B + b^3*B + 3
*a^3*C + 3*a*b^2*C)*EllipticF[(c + d*x)/2, 2])/(3*d) - (2*b*(6*a^2*B - b^2*B - 3*a*b*C)*Sqrt[Cos[c + d*x]]*Sin
[c + d*x])/(3*d) - (2*b^2*(5*a*B - b*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*B*(a + b*Cos[c + d*x])^2
*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])
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Rubi [A] time = 0.558461, antiderivative size = 202, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{3029, 2989, 3033, 3023, 2748, 2641, 2639} \[ \frac{2 \left (9 a^2 b B+3 a^3 C+3 a b^2 C+b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (-15 a^2 b C+5 a^3 B-15 a b^2 B-3 b^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}-\frac{2 b \left (6 a^2 B-3 a b C-b^2 B\right ) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}-\frac{2 b^2 (5 a B-b C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d}+\frac{2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{d \sqrt{\cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
[Out]
(-2*(5*a^3*B - 15*a*b^2*B - 15*a^2*b*C - 3*b^3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(9*a^2*b*B + b^3*B + 3
*a^3*C + 3*a*b^2*C)*EllipticF[(c + d*x)/2, 2])/(3*d) - (2*b*(6*a^2*B - b^2*B - 3*a*b*C)*Sqrt[Cos[c + d*x]]*Sin
[c + d*x])/(3*d) - (2*b^2*(5*a*B - b*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + (2*a*B*(a + b*Cos[c + d*x])^2
*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])
Rule 3029
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
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 3033
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[(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*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[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 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\int \frac{(a+b \cos (c+d x))^3 (B+C \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+2 \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{2} a (5 b B+a C)-\frac{1}{2} \left (a^2 B-b^2 B-2 a b C\right ) \cos (c+d x)-\frac{1}{2} b (5 a B-b C) \cos ^2(c+d x)\right )}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b^2 (5 a B-b C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{4}{5} \int \frac{\frac{5}{4} a^2 (5 b B+a C)-\frac{1}{4} \left (5 a^3 B-15 a b^2 B-15 a^2 b C-3 b^3 C\right ) \cos (c+d x)-\frac{5}{4} b \left (6 a^2 B-b^2 B-3 a b C\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (6 a^2 B-b^2 B-3 a b C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 a B-b C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{8}{15} \int \frac{\frac{5}{8} \left (9 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right )-\frac{3}{8} \left (5 a^3 B-15 a b^2 B-15 a^2 b C-3 b^3 C\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b \left (6 a^2 B-b^2 B-3 a b C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 a B-b C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{1}{3} \left (9 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (-5 a^3 B+15 a b^2 B+15 a^2 b C+3 b^3 C\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (5 a^3 B-15 a b^2 B-15 a^2 b C-3 b^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (9 a^2 b B+b^3 B+3 a^3 C+3 a b^2 C\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 b \left (6 a^2 B-b^2 B-3 a b C\right ) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-\frac{2 b^2 (5 a B-b C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.11544, size = 150, normalized size = 0.74 \[ \frac{10 \left (9 a^2 b B+3 a^3 C+3 a b^2 C+b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\left (90 a^2 b C-30 a^3 B+90 a b^2 B+18 b^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\frac{\sin (c+d x) \left (3 \left (10 a^3 B+b^3 C \cos (2 (c+d x))+b^3 C\right )+10 b^2 (3 a C+b B) \cos (c+d x)\right )}{\sqrt{\cos (c+d x)}}}{15 d} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
[Out]
((-30*a^3*B + 90*a*b^2*B + 90*a^2*b*C + 18*b^3*C)*EllipticE[(c + d*x)/2, 2] + 10*(9*a^2*b*B + b^3*B + 3*a^3*C
+ 3*a*b^2*C)*EllipticF[(c + d*x)/2, 2] + ((10*b^2*(b*B + 3*a*C)*Cos[c + d*x] + 3*(10*a^3*B + b^3*C + b^3*C*Cos
[2*(c + d*x)]))*Sin[c + d*x])/Sqrt[Cos[c + d*x]])/(15*d)
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Maple [B] time = 0.718, size = 867, normalized size = 4.3 \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*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)
[Out]
-2/15*(-24*C*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
4*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(5*B*b+15*C*a+6*C*b)*sin(1/2*d*x+1/2*c)^4*cos(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)*(15*B*a^3+5*B*b^3+15*C*a*b^2+3*C*b^3)*sin(1/2
*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+45*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)*Elli
pticF(cos(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)+5*b^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)*EllipticF(cos(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)+15*B*(-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)^(
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^3-45*B*(-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)^(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+15*a^3*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)*EllipticF(cos(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)+15*C*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*
d*x+1/2*c)^2)^(1/2)-45*C*(-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)^(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-9*C*(-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)^(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*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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\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*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(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{C b^{3} \cos \left (d x + c\right )^{4} + B a^{3} +{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (C a^{2} b + B a b^{2}\right )} \cos \left (d x + c\right )^{2} +{\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{\frac{3}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
integral((C*b^3*cos(d*x + c)^4 + B*a^3 + (3*C*a*b^2 + B*b^3)*cos(d*x + c)^3 + 3*(C*a^2*b + B*a*b^2)*cos(d*x +
c)^2 + (C*a^3 + 3*B*a^2*b)*cos(d*x + c))/cos(d*x + c)^(3/2), 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*(B*cos(d*x+c)+C*cos(d*x+c)**2)/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 (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\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*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^3/cos(d*x + c)^(5/2), x)