3.233 \(\int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=131 \[ \frac {6 \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{5 d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d} \]

[Out]

-2/5*(b*cos(d*x+c)-a*sin(d*x+c))*(a*cos(d*x+c)+b*sin(d*x+c))^(3/2)/d+6/5*(a^2+b^2)*(cos(1/2*c+1/2*d*x-1/2*arct
an(a,b))^2)^(1/2)/cos(1/2*c+1/2*d*x-1/2*arctan(a,b))*EllipticE(sin(1/2*c+1/2*d*x-1/2*arctan(a,b)),2^(1/2))*(a*
cos(d*x+c)+b*sin(d*x+c))^(1/2)/d/((a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^(1/2))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3073, 3078, 2639} \[ \frac {6 \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{5 d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(3/2))/(5*d) + (6*(a^2 + b^2)*Elliptic
E[(c + d*x - ArcTan[a, b])/2, 2]*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]])/(5*d*Sqrt[(a*Cos[c + d*x] + b*Sin[c +
d*x])/Sqrt[a^2 + b^2]])

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 3073

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*Cos[c + d*x]
- a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n - 1))/(d*n), x] + Dist[((n - 1)*(a^2 + b^2))/n, Int[(a*
Cos[c + d*x] + b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[(n
- 1)/2] && GtQ[n, 1]

Rule 3078

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[c + d*x] +
b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2])^n, Int[Cos[c + d*x - ArcTan[a, b]]^n, x]
, x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])

Rubi steps

\begin {align*} \int (a \cos (c+d x)+b \sin (c+d x))^{5/2} \, dx &=-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}+\frac {1}{5} \left (3 \left (a^2+b^2\right )\right ) \int \sqrt {a \cos (c+d x)+b \sin (c+d x)} \, dx\\ &=-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}+\frac {\left (3 \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}\right ) \int \sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )} \, dx}{5 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}\\ &=-\frac {2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^{3/2}}{5 d}+\frac {6 \left (a^2+b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{5 d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}\\ \end {align*}

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Mathematica [C]  time = 1.60, size = 256, normalized size = 1.95 \[ \frac {\sqrt {a \cos (c+d x)+b \sin (c+d x)} \left (b \left (a^2-b^2\right ) \sin (2 (c+d x))+6 a \left (a^2+b^2\right )-2 a b^2 \cos (2 (c+d x))\right )-\frac {3 \left (a^2+b^2\right )^2 \cos \left (-\tan ^{-1}\left (\frac {b}{a}\right )+c+d x\right ) \left (b \sin \left (-\tan ^{-1}\left (\frac {b}{a}\right )+c+d x\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\tan ^{-1}\left (\frac {b}{a}\right )\right )\right )+\sqrt {\sin ^2\left (-\tan ^{-1}\left (\frac {b}{a}\right )+c+d x\right )} \left (2 a \cos \left (-\tan ^{-1}\left (\frac {b}{a}\right )+c+d x\right )-b \sin \left (-\tan ^{-1}\left (\frac {b}{a}\right )+c+d x\right )\right )\right )}{\sqrt {\sin ^2\left (-\tan ^{-1}\left (\frac {b}{a}\right )+c+d x\right )} \left (a \sqrt {\frac {b^2}{a^2}+1} \cos \left (-\tan ^{-1}\left (\frac {b}{a}\right )+c+d x\right )\right )^{3/2}}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]]*(6*a*(a^2 + b^2) - 2*a*b^2*Cos[2*(c + d*x)] + b*(a^2 - b^2)*Sin[2*(c +
d*x)]) - (3*(a^2 + b^2)^2*Cos[c + d*x - ArcTan[b/a]]*(b*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[c + d*x - A
rcTan[b/a]]^2]*Sin[c + d*x - ArcTan[b/a]] + Sqrt[Sin[c + d*x - ArcTan[b/a]]^2]*(2*a*Cos[c + d*x - ArcTan[b/a]]
- b*Sin[c + d*x - ArcTan[b/a]])))/((a*Sqrt[1 + b^2/a^2]*Cos[c + d*x - ArcTan[b/a]])^(3/2)*Sqrt[Sin[c + d*x -
ArcTan[b/a]]^2]))/(5*b*d)

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)*sqrt(a*cos(d*x + c) + b*sin(d*x
+ c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.37, size = 246, normalized size = 1.88 \[ -\frac {\left (a^{2}+b^{2}\right )^{\frac {3}{2}} \left (6 \sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \sqrt {-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \EllipticE \left (\sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \sqrt {-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (d x +c -\arctan \left (-a , b\right )\right )\right )+2 \left (\sin ^{2}\left (d x +c -\arctan \left (-a , b\right )\right )\right )\right )}{5 \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(a^2+b^2)^(3/2)*(6*(1+sin(d*x+c-arctan(-a,b)))^(1/2)*(-2*sin(d*x+c-arctan(-a,b))+2)^(1/2)*(-sin(d*x+c-arc
tan(-a,b)))^(1/2)*EllipticE((1+sin(d*x+c-arctan(-a,b)))^(1/2),1/2*2^(1/2))-3*(1+sin(d*x+c-arctan(-a,b)))^(1/2)
*(-2*sin(d*x+c-arctan(-a,b))+2)^(1/2)*(-sin(d*x+c-arctan(-a,b)))^(1/2)*EllipticF((1+sin(d*x+c-arctan(-a,b)))^(
1/2),1/2*2^(1/2))-2*sin(d*x+c-arctan(-a,b))^4+2*sin(d*x+c-arctan(-a,b))^2)/cos(d*x+c-arctan(-a,b))/(sin(d*x+c-
arctan(-a,b))*(a^2+b^2)^(1/2))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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