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*(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]])

________________________________________________________________________________________

Rubi [A]  time = 0.057692, antiderivative size = 131, normalized size of antiderivative = 1., 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 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])

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 (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*}

Mathematica [C]  time = 1.61262, size = 256, normalized size = 1.95 \[ \frac{\sqrt{a \cos (c+d x)+b \sin (c+d x)} \left (b \left (a^2-b^2\right ) \sin (2 (c+d x))+6 a \left (a^2+b^2\right )-2 a b^2 \cos (2 (c+d x))\right )-\frac{3 \left (a^2+b^2\right )^2 \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \left (b \sin \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )+\sqrt{\sin ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )} \left (2 a \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )-b \sin \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )\right )}{\sqrt{\sin ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )} \left (a \sqrt{\frac{b^2}{a^2}+1} \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )^{3/2}}}{5 b d} \]

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)

________________________________________________________________________________________

Maple [A]  time = 1.464, size = 246, normalized size = 1.9 \begin{align*} -{\frac{1}{5\,\cos \left ( dx+c-\arctan \left ( -a,b \right ) \right ) d} \left ({a}^{2}+{b}^{2} \right ) ^{{\frac{3}{2}}} \left ( 6\,\sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) +2}\sqrt{-\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }{\it EllipticE} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },1/2\,\sqrt{2} \right ) -3\,\sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) +2}\sqrt{-\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },1/2\,\sqrt{2} \right ) -2\, \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{4}+2\, \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \sqrt{{a}^{2}+{b}^{2}}}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac{5}{2}}\,{d x} \end{align*}

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

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)