3.568 \(\int \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx\)

Optimal. Leaf size=160 \[ \frac{2 \left (9 a^2+7 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 \left (9 a^2+7 b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{20 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a b \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{20 a b \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 b^2 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{9 d} \]

[Out]

(2*(9*a^2 + 7*b^2)*EllipticE[(c + d*x)/2, 2])/(15*d) + (20*a*b*EllipticF[(c + d*x)/2, 2])/(21*d) + (20*a*b*Sqr
t[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*(9*a^2 + 7*b^2)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (4*a*b*Cos
[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + (2*b^2*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d)

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Rubi [A]  time = 0.120122, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{2 \left (9 a^2+7 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 \left (9 a^2+7 b^2\right ) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{20 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{4 a b \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{20 a b \sin (c+d x) \sqrt{\cos (c+d x)}}{21 d}+\frac{2 b^2 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2,x]

[Out]

(2*(9*a^2 + 7*b^2)*EllipticE[(c + d*x)/2, 2])/(15*d) + (20*a*b*EllipticF[(c + d*x)/2, 2])/(21*d) + (20*a*b*Sqr
t[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*(9*a^2 + 7*b^2)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (4*a*b*Cos
[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + (2*b^2*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*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 \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2 \, dx &=(2 a b) \int \cos ^{\frac{7}{2}}(c+d x) \, dx+\int \cos ^{\frac{5}{2}}(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac{4 a b \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 b^2 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{7} (10 a b) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{9} \left (9 a^2+7 b^2\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{20 a b \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 \left (9 a^2+7 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{4 a b \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 b^2 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{21} (10 a b) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{15} \left (9 a^2+7 b^2\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 \left (9 a^2+7 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{20 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}+\frac{20 a b \sqrt{\cos (c+d x)} \sin (c+d x)}{21 d}+\frac{2 \left (9 a^2+7 b^2\right ) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{4 a b \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 b^2 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.754149, size = 113, normalized size = 0.71 \[ \frac{84 \left (9 a^2+7 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt{\cos (c+d x)} \left (7 \left (36 a^2+43 b^2\right ) \cos (c+d x)+5 b (36 a \cos (2 (c+d x))+156 a+7 b \cos (3 (c+d x)))\right )+600 a b F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{630 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(5/2)*(a + b*Cos[c + d*x])^2,x]

[Out]

(84*(9*a^2 + 7*b^2)*EllipticE[(c + d*x)/2, 2] + 600*a*b*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(7*(36*
a^2 + 43*b^2)*Cos[c + d*x] + 5*b*(156*a + 36*a*Cos[2*(c + d*x)] + 7*b*Cos[3*(c + d*x)]))*Sin[c + d*x])/(630*d)

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Maple [B]  time = 2.832, size = 398, normalized size = 2.5 \begin{align*} -{\frac{2}{315\,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 ( -1120\,{b}^{2} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{10}\cos \left ( 1/2\,dx+c/2 \right ) + \left ( 1440\,ab+2240\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{8}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -504\,{a}^{2}-2160\,ab-2072\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{6}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( 504\,{a}^{2}+1680\,ab+952\,{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 ( -126\,{a}^{2}-480\,ab-168\,{b}^{2} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +150\,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 ) -189\,\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}-147\,\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)^(5/2)*(a+b*cos(d*x+c))^2,x)

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*b^2*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/
2*c)+(1440*a*b+2240*b^2)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-504*a^2-2160*a*b-2072*b^2)*sin(1/2*d*x+1/2*
c)^6*cos(1/2*d*x+1/2*c)+(504*a^2+1680*a*b+952*b^2)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-126*a^2-480*a*b-1
68*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+150*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))-189*(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^2-147*(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*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{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \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(cos(d*x+c)^(5/2)*(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^2*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 ({\left (b^{2} \cos \left (d x + c\right )^{4} + 2 \, a b \cos \left (d x + c\right )^{3} + a^{2} \cos \left (d x + c\right )^{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)^(5/2)*(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((b^2*cos(d*x + c)^4 + 2*a*b*cos(d*x + c)^3 + a^2*cos(d*x + c)^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)**(5/2)*(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \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(cos(d*x+c)^(5/2)*(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^2*cos(d*x + c)^(5/2), x)