3.679 \(\int \frac{(a+b \cos (c+d x)) (A+C \cos ^2(c+d x))}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=95 \[ \frac{2 a (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 A b \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]
[Out]
(-2*b*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*A*Sin[c +
d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*A*b*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])
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Rubi [A] time = 0.185646, antiderivative size = 95, 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 =
{3032, 3021, 2748, 2641, 2639} \[ \frac{2 a (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 A b \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
[Out]
(-2*b*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (2*a*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*A*Sin[c +
d*x])/(3*d*Cos[c + d*x]^(3/2)) + (2*A*b*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])
Rule 3032
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 + 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)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((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, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[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)) \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 a A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{\frac{3 A b}{2}+\frac{1}{2} a (A+3 C) \cos (c+d x)+\frac{3}{2} b C \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A b \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{4}{3} \int \frac{\frac{1}{4} a (A+3 C)-\frac{3}{4} b (A-C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A b \sin (c+d x)}{d \sqrt{\cos (c+d x)}}-(b (A-C)) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} (a (A+3 C)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 b (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a A \sin (c+d x)}{3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 A b \sin (c+d x)}{d \sqrt{\cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.759188, size = 76, normalized size = 0.8 \[ \frac{\frac{2 A \sin (c+d x) (a+3 b \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)}+2 a (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+6 b (C-A) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(5/2),x]
[Out]
(6*b*(-A + C)*EllipticE[(c + d*x)/2, 2] + 2*a*(A + 3*C)*EllipticF[(c + d*x)/2, 2] + (2*A*(a + 3*b*Cos[c + d*x]
)*Sin[c + d*x])/Cos[c + d*x]^(3/2))/(3*d)
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Maple [B] time = 0.891, size = 614, normalized size = 6.5 \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))*(A+C*cos(d*x+c)^2)/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*(2*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))*a*sin(1/2*d*x+1/2*c)^2+6*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b*sin(1/2*d*x+1/2*c)^2-12*A*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*C*(
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*sin(1/2*d
*x+1/2*c)^2-6*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(
1/2))*b*sin(1/2*d*x+1/2*c)^2-a*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)*EllipticF(cos(1
/2*d*x+1/2*c),2^(1/2))-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+2*A*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+6*A*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2
-3*a*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))+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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)*(-2*si
n(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 (C \cos \left (d x + c\right )^{2} + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}}{\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))*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)/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 \cos \left (d x + c\right )^{3} + C a \cos \left (d x + c\right )^{2} + A b \cos \left (d x + c\right ) + A a}{\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))*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
integral((C*b*cos(d*x + c)^3 + C*a*cos(d*x + c)^2 + A*b*cos(d*x + c) + A*a)/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))*(A+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} + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}}{\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))*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)/cos(d*x + c)^(5/2), x)