3.678 \(\int \frac{(a+b \cos (c+d x)) (A+C \cos ^2(c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=95 \[ -\frac{2 a (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b C \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

[Out]

(-2*a*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (2*b*(3*A + C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*A*Sin[c +
d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*b*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.173419, 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, 3023, 2748, 2641, 2639} \[ -\frac{2 a (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 b C \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

(-2*a*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (2*b*(3*A + C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a*A*Sin[c +
d*x])/(d*Sqrt[Cos[c + d*x]]) + (2*b*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)

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 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)) \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+2 \int \frac{\frac{A b}{2}-\frac{1}{2} a (A-C) \cos (c+d x)+\frac{1}{2} b C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b C \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{4}{3} \int \frac{\frac{1}{4} b (3 A+C)-\frac{3}{4} a (A-C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b C \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}-(a (A-C)) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} (b (3 A+C)) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{2 a (A-C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a A \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 b C \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.468352, size = 78, normalized size = 0.82 \[ \frac{\frac{2 \sin (c+d x) (3 a A+b C \cos (c+d x))}{\sqrt{\cos (c+d x)}}+6 a (C-A) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 b (3 A+C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

(6*a*(-A + C)*EllipticE[(c + d*x)/2, 2] + 2*b*(3*A + C)*EllipticF[(c + d*x)/2, 2] + (2*(3*a*A + b*C*Cos[c + d*
x])*Sin[c + d*x])/Sqrt[Cos[c + d*x]])/(3*d)

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Maple [B]  time = 0.388, size = 294, normalized size = 3.1 \begin{align*} -{\frac{2}{3\,d} \left ( 4\,Cb\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+3\,Ab\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +3\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a-6\,Aa\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+Cb\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -3\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a-2\,Cb\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \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)^(3/2),x)

[Out]

-2/3*(4*C*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+3*A*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))+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))*a-6*A*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+C*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))-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))*a-2*C*b*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*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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c) + a)/cos(d*x + c)^(3/2), 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{3}{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)^(3/2),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)^(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))*(A+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2),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{3}{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)^(3/2),x, algorithm="giac")

[Out]

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