3.873 \(\int \frac{(a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac{9}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=204 \[ \frac{2 \left (3 a^2 b B+a^3 C+9 a b^2 C+3 b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (15 a^2 b C+3 a^3 B+15 a b^2 B-5 b^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a \left (3 a^2 B+15 a b C+14 b^2 B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 (5 a C+9 b B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
[Out]
(-2*(3*a^3*B + 15*a*b^2*B + 15*a^2*b*C - 5*b^3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(3*a^2*b*B + 3*b^3*B +
a^3*C + 9*a*b^2*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(9*b*B + 5*a*C)*Sin[c + d*x])/(15*d*Cos[c + d*x]
^(3/2)) + (2*a*(3*a^2*B + 14*b^2*B + 15*a*b*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a*B*(a + b*Cos[c +
d*x])^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
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Rubi [A] time = 0.546613, antiderivative size = 204, 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, 3031, 3021, 2748, 2641, 2639} \[ \frac{2 \left (3 a^2 b B+a^3 C+9 a b^2 C+3 b^3 B\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{2 \left (15 a^2 b C+3 a^3 B+15 a b^2 B-5 b^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a \left (3 a^2 B+15 a b C+14 b^2 B\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a^2 (5 a C+9 b B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a B \sin (c+d x) (a+b \cos (c+d x))^2}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
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]^(9/2),x]
[Out]
(-2*(3*a^3*B + 15*a*b^2*B + 15*a^2*b*C - 5*b^3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*(3*a^2*b*B + 3*b^3*B +
a^3*C + 9*a*b^2*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^2*(9*b*B + 5*a*C)*Sin[c + d*x])/(15*d*Cos[c + d*x]
^(3/2)) + (2*a*(3*a^2*B + 14*b^2*B + 15*a*b*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]]) + (2*a*B*(a + b*Cos[c +
d*x])^2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
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 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 3021
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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{9}{2}}(c+d x)} \, dx &=\int \frac{(a+b \cos (c+d x))^3 (B+C \cos (c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \cos (c+d x)) \left (\frac{1}{2} a (9 b B+5 a C)+\frac{1}{2} \left (3 a^2 B+5 b^2 B+10 a b C\right ) \cos (c+d x)-\frac{1}{2} b (a B-5 b C) \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{4}{15} \int \frac{-\frac{3}{4} a \left (3 a^2 B+14 b^2 B+15 a b C\right )-\frac{5}{4} \left (3 a^2 b B+3 b^3 B+a^3 C+9 a b^2 C\right ) \cos (c+d x)+\frac{3}{4} b^2 (a B-5 b C) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (3 a^2 B+14 b^2 B+15 a b C\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{8}{15} \int \frac{-\frac{5}{8} \left (3 a^2 b B+3 b^3 B+a^3 C+9 a b^2 C\right )+\frac{3}{8} \left (3 a^3 B+15 a b^2 B+15 a^2 b C-5 b^3 C\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (3 a^2 B+14 b^2 B+15 a b C\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{1}{3} \left (-3 a^2 b B-3 b^3 B-a^3 C-9 a b^2 C\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (3 a^3 B+15 a b^2 B+15 a^2 b C-5 b^3 C\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 \left (3 a^3 B+15 a b^2 B+15 a^2 b C-5 b^3 C\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 \left (3 a^2 b B+3 b^3 B+a^3 C+9 a b^2 C\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 (9 b B+5 a C) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 a \left (3 a^2 B+14 b^2 B+15 a b C\right ) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 a B (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 2.02626, size = 176, normalized size = 0.86 \[ \frac{9 a \left (a^2 B+5 a b C+5 b^2 B\right ) \sin (2 (c+d x))+10 \left (3 a^2 b B+a^3 C+9 a b^2 C+3 b^3 B\right ) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-6 \left (15 a^2 b C+3 a^3 B+15 a b^2 B-5 b^3 C\right ) \cos ^{\frac{3}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+10 a^2 (a C+3 b B) \sin (c+d x)+6 a^3 B \tan (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)} \]
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]^(9/2),x]
[Out]
(-6*(3*a^3*B + 15*a*b^2*B + 15*a^2*b*C - 5*b^3*C)*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] + 10*(3*a^2*b*B
+ 3*b^3*B + a^3*C + 9*a*b^2*C)*Cos[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] + 10*a^2*(3*b*B + a*C)*Sin[c + d*
x] + 9*a*(a^2*B + 5*b^2*B + 5*a*b*C)*Sin[2*(c + d*x)] + 6*a^3*B*Tan[c + d*x])/(15*d*Cos[c + d*x]^(3/2))
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Maple [B] time = 2.143, size = 997, normalized size = 4.9 \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)^(9/2),x)
[Out]
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d
*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*b^3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/
2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*C*a*b^2*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2*C*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2*a^2*(3*B*b+C*a)*(-
1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^
(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+6*a*b*(B*b+C*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)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)
^2/(2*sin(1/2*d*x+1/2*c)^2-1)-2/5*a^3*B/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2
-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(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)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(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)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*
x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(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)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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*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{9}{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)^(9/2),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)^(9/2), 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{7}{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)^(9/2),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)^(7/2), 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)**(9/2),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{9}{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)^(9/2),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)^(9/2), x)