3.570 \(\int \sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2 \, dx\)

Optimal. Leaf size=101 \[ \frac{2 \left (5 a^2+3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a b \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 b^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

[Out]

(2*(5*a^2 + 3*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a*b*EllipticF[(c + d*x)/2, 2])/(3*d) + (4*a*b*Sqrt[Co
s[c + d*x]]*Sin[c + d*x])/(3*d) + (2*b^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.0904047, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2789, 2635, 2641, 3014, 2639} \[ \frac{2 \left (5 a^2+3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a b \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 b^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2,x]

[Out]

(2*(5*a^2 + 3*b^2)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a*b*EllipticF[(c + d*x)/2, 2])/(3*d) + (4*a*b*Sqrt[Co
s[c + d*x]]*Sin[c + d*x])/(3*d) + (2*b^2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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 \sqrt{\cos (c+d x)} (a+b \cos (c+d x))^2 \, dx &=(2 a b) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\int \sqrt{\cos (c+d x)} \left (a^2+b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac{4 a b \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 b^2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{1}{3} (2 a b) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (5 a^2+3 b^2\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (5 a^2+3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a b \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 b^2 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.298858, size = 79, normalized size = 0.78 \[ \frac{6 \left (5 a^2+3 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+20 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 b \sin (c+d x) \sqrt{\cos (c+d x)} (10 a+3 b \cos (c+d x))}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])^2,x]

[Out]

(6*(5*a^2 + 3*b^2)*EllipticE[(c + d*x)/2, 2] + 20*a*b*EllipticF[(c + d*x)/2, 2] + 2*b*Sqrt[Cos[c + d*x]]*(10*a
+ 3*b*Cos[c + d*x])*Sin[c + d*x])/(15*d)

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Maple [B]  time = 2.661, size = 321, normalized size = 3.2 \begin{align*} -{\frac{2}{15\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( -24\,{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 40\,ab+24\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -20\,ab-6\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +10\,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 ) -15\,\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}-9\,\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 ){b}^{2} \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \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(cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))^2,x)

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+(40*a*b+24*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-20*a*b-6*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*
c)+10*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))-
15*(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^2-9*(
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)/(-2*si
n(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{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^2*sqrt(cos(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(cos(d*x + c)), 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(cos(d*x+c)**(1/2)*(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [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(cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out