3.168 \(\int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^4 \, dx\)

Optimal. Leaf size=147 \[ \frac{32 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{152 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^4 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{9 d}+\frac{8 a^4 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{122 a^4 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{32 a^4 \sin (c+d x) \sqrt{\cos (c+d x)}}{7 d} \]

[Out]

(152*a^4*EllipticE[(c + d*x)/2, 2])/(15*d) + (32*a^4*EllipticF[(c + d*x)/2, 2])/(7*d) + (32*a^4*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(7*d) + (122*a^4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (8*a^4*Cos[c + d*x]^(5/2)*Sin[c
+ d*x])/(7*d) + (2*a^4*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d)

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Rubi [A]  time = 0.163782, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2757, 2639, 2635, 2641} \[ \frac{32 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{152 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{2 a^4 \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x)}{9 d}+\frac{8 a^4 \sin (c+d x) \cos ^{\frac{5}{2}}(c+d x)}{7 d}+\frac{122 a^4 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{45 d}+\frac{32 a^4 \sin (c+d x) \sqrt{\cos (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^4,x]

[Out]

(152*a^4*EllipticE[(c + d*x)/2, 2])/(15*d) + (32*a^4*EllipticF[(c + d*x)/2, 2])/(7*d) + (32*a^4*Sqrt[Cos[c + d
*x]]*Sin[c + d*x])/(7*d) + (122*a^4*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(45*d) + (8*a^4*Cos[c + d*x]^(5/2)*Sin[c
+ d*x])/(7*d) + (2*a^4*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(9*d)

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

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]

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]

Rubi steps

\begin{align*} \int \sqrt{\cos (c+d x)} (a+a \cos (c+d x))^4 \, dx &=\int \left (a^4 \sqrt{\cos (c+d x)}+4 a^4 \cos ^{\frac{3}{2}}(c+d x)+6 a^4 \cos ^{\frac{5}{2}}(c+d x)+4 a^4 \cos ^{\frac{7}{2}}(c+d x)+a^4 \cos ^{\frac{9}{2}}(c+d x)\right ) \, dx\\ &=a^4 \int \sqrt{\cos (c+d x)} \, dx+a^4 \int \cos ^{\frac{9}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^{\frac{7}{2}}(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx\\ &=\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{8 a^4 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{12 a^4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{8 a^4 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{9} \left (7 a^4\right ) \int \cos ^{\frac{5}{2}}(c+d x) \, dx+\frac{1}{3} \left (4 a^4\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{7} \left (20 a^4\right ) \int \cos ^{\frac{3}{2}}(c+d x) \, dx+\frac{1}{5} \left (18 a^4\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{46 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{8 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{32 a^4 \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d}+\frac{122 a^4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{8 a^4 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}+\frac{1}{15} \left (7 a^4\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{21} \left (20 a^4\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{152 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d}+\frac{32 a^4 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 d}+\frac{32 a^4 \sqrt{\cos (c+d x)} \sin (c+d x)}{7 d}+\frac{122 a^4 \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac{8 a^4 \cos ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{2 a^4 \cos ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{9 d}\\ \end{align*}

Mathematica [C]  time = 6.14535, size = 532, normalized size = 3.62 \[ -\frac{19 \csc (c) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^4 \left (\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )}{\sqrt{\tan ^2(c)+1} \sqrt{1-\cos \left (\tan ^{-1}(\tan (c))+d x\right )} \sqrt{\cos \left (\tan ^{-1}(\tan (c))+d x\right )+1} \sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}-\frac{\frac{\tan (c) \sin \left (\tan ^{-1}(\tan (c))+d x\right )}{\sqrt{\tan ^2(c)+1}}+\frac{2 \cos ^2(c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}{\sin ^2(c)+\cos ^2(c)}}{\sqrt{\cos (c) \sqrt{\tan ^2(c)+1} \cos \left (\tan ^{-1}(\tan (c))+d x\right )}}\right )}{60 d}-\frac{2 \csc (c) \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^4 \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin (c) \left (-\sqrt{\cot ^2(c)+1}\right ) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )}{7 d \sqrt{\cot ^2(c)+1}}+\sqrt{\cos (c+d x)} \sec ^8\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \cos (c+d x)+a)^4 \left (\frac{17 \sin (c) \cos (d x)}{56 d}+\frac{127 \sin (2 c) \cos (2 d x)}{1440 d}+\frac{\sin (3 c) \cos (3 d x)}{56 d}+\frac{\sin (4 c) \cos (4 d x)}{576 d}+\frac{17 \cos (c) \sin (d x)}{56 d}+\frac{127 \cos (2 c) \sin (2 d x)}{1440 d}+\frac{\cos (3 c) \sin (3 d x)}{56 d}+\frac{\cos (4 c) \sin (4 d x)}{576 d}-\frac{19 \cot (c)}{30 d}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^4,x]

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^4*Sec[c/2 + (d*x)/2]^8*((-19*Cot[c])/(30*d) + (17*Cos[d*x]*Sin[c])/(56
*d) + (127*Cos[2*d*x]*Sin[2*c])/(1440*d) + (Cos[3*d*x]*Sin[3*c])/(56*d) + (Cos[4*d*x]*Sin[4*c])/(576*d) + (17*
Cos[c]*Sin[d*x])/(56*d) + (127*Cos[2*c]*Sin[2*d*x])/(1440*d) + (Cos[3*c]*Sin[3*d*x])/(56*d) + (Cos[4*c]*Sin[4*
d*x])/(576*d)) - (2*(a + a*Cos[c + d*x])^4*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]
]]^2]*Sec[c/2 + (d*x)/2]^8*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c
]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(7*d*Sqrt[1 + Cot[c]^2]) - (19*(a
+ a*Cos[c + d*x])^4*Csc[c]*Sec[c/2 + (d*x)/2]^8*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan
[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[
c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c
]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^
2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(60*d)

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Maple [A]  time = 2.234, size = 260, normalized size = 1.8 \begin{align*} -{\frac{8\,{a}^{4}}{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 ( 280\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}-120\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}+34\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}+72\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-485\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+180\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -399\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +219\,\cos \left ( 1/2\,dx+c/2 \right ) \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+cos(d*x+c)*a)^4,x)

[Out]

-8/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*(280*cos(1/2*d*x+1/2*c)^11-120*cos(1/2*d*x+
1/2*c)^9+34*cos(1/2*d*x+1/2*c)^7+72*cos(1/2*d*x+1/2*c)^5-485*cos(1/2*d*x+1/2*c)^3+180*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-399*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+219*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{\left (a \cos \left (d x + c\right ) + a\right )}^{4} \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+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4*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 (a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}\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+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((a^4*cos(d*x + c)^4 + 4*a^4*cos(d*x + c)^3 + 6*a^4*cos(d*x + c)^2 + 4*a^4*cos(d*x + c) + a^4)*sqrt(co
s(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+a*cos(d*x+c))**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + a\right )}^{4} \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+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4*sqrt(cos(d*x + c)), x)