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

Optimal. Leaf size=265 \[ -\frac{2 \sqrt{2} a \left (4 a^2-b^2\right ) \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}} \text{EllipticF}\left (c+d x-\frac{\pi }{4},\frac{2 b}{2 a+b}\right )}{15 d \sqrt{2 a+b \sin (2 c+2 d x)}}+\frac{\left (92 a^2+9 b^2\right ) \sqrt{2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{60 \sqrt{2} d \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}-\frac{2 \sqrt{2} a b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{15 d} \]

[Out]

(-2*Sqrt[2]*a*b*Cos[2*c + 2*d*x]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(15*d) - (b*Cos[2*c + 2*d*x]*(2*a + b*Sin[2*c
+ 2*d*x])^(3/2))/(20*Sqrt[2]*d) + ((92*a^2 + 9*b^2)*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[2*a + b*S
in[2*c + 2*d*x]])/(60*Sqrt[2]*d*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)]) - (2*Sqrt[2]*a*(4*a^2 - b^2)*Ellip
ticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(15*d*Sqrt[2*a + b*Sin[2*c +
2*d*x]])

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Rubi [A]  time = 0.365848, antiderivative size = 265, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {2666, 2656, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \sqrt{2} a \left (4 a^2-b^2\right ) \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}} F\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{15 d \sqrt{2 a+b \sin (2 c+2 d x)}}+\frac{\left (92 a^2+9 b^2\right ) \sqrt{2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )}{60 \sqrt{2} d \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}-\frac{2 \sqrt{2} a b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a*b*Cos[2*c + 2*d*x]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(15*d) - (b*Cos[2*c + 2*d*x]*(2*a + b*Sin[2*c
+ 2*d*x])^(3/2))/(20*Sqrt[2]*d) + ((92*a^2 + 9*b^2)*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[2*a + b*S
in[2*c + 2*d*x]])/(60*Sqrt[2]*d*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)]) - (2*Sqrt[2]*a*(4*a^2 - b^2)*Ellip
ticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(15*d*Sqrt[2*a + b*Sin[2*c +
2*d*x]])

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2656

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

Rule 2753

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

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2663

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

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

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

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x) \sin (c+d x))^{5/2} \, dx &=\int \left (a+\frac{1}{2} b \sin (2 c+2 d x)\right )^{5/2} \, dx\\ &=-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{2}{5} \int \sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)} \left (\frac{1}{8} \left (20 a^2+3 b^2\right )+2 a b \sin (2 c+2 d x)\right ) \, dx\\ &=-\frac{2 \sqrt{2} a b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{15 d}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{4}{15} \int \frac{\frac{1}{16} a \left (60 a^2+17 b^2\right )+\frac{1}{32} b \left (92 a^2+9 b^2\right ) \sin (2 c+2 d x)}{\sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}} \, dx\\ &=-\frac{2 \sqrt{2} a b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{15 d}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}-\frac{1}{15} \left (2 a \left (4 a^2-b^2\right )\right ) \int \frac{1}{\sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}} \, dx+\frac{1}{60} \left (92 a^2+9 b^2\right ) \int \sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)} \, dx\\ &=-\frac{2 \sqrt{2} a b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{15 d}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{\left (\left (92 a^2+9 b^2\right ) \sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}\right ) \int \sqrt{\frac{a}{a+\frac{b}{2}}+\frac{b \sin (2 c+2 d x)}{2 \left (a+\frac{b}{2}\right )}} \, dx}{60 \sqrt{\frac{a+\frac{1}{2} b \sin (2 c+2 d x)}{a+\frac{b}{2}}}}-\frac{\left (2 a \left (4 a^2-b^2\right ) \sqrt{\frac{a+\frac{1}{2} b \sin (2 c+2 d x)}{a+\frac{b}{2}}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+\frac{b}{2}}+\frac{b \sin (2 c+2 d x)}{2 \left (a+\frac{b}{2}\right )}}} \, dx}{15 \sqrt{a+\frac{1}{2} b \sin (2 c+2 d x)}}\\ &=-\frac{2 \sqrt{2} a b \cos (2 c+2 d x) \sqrt{2 a+b \sin (2 c+2 d x)}}{15 d}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^{3/2}}{20 \sqrt{2} d}+\frac{\left (92 a^2+9 b^2\right ) E\left (c-\frac{\pi }{4}+d x|\frac{2 b}{2 a+b}\right ) \sqrt{2 a+b \sin (2 c+2 d x)}}{60 \sqrt{2} d \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac{2 \sqrt{2} a \left (4 a^2-b^2\right ) F\left (c-\frac{\pi }{4}+d x|\frac{2 b}{2 a+b}\right ) \sqrt{\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}}}{15 d \sqrt{2 a+b \sin (2 c+2 d x)}}\\ \end{align*}

Mathematica [A]  time = 1.91393, size = 202, normalized size = 0.76 \[ \frac{-32 a \left (4 a^2-b^2\right ) \sqrt{\frac{2 a+b \sin (2 (c+d x))}{2 a+b}} \text{EllipticF}\left (c+d x-\frac{\pi }{4},\frac{2 b}{2 a+b}\right )+2 \left (92 a^2 b+184 a^3+18 a b^2+9 b^3\right ) \sqrt{\frac{2 a+b \sin (2 (c+d x))}{2 a+b}} E\left (c+d x-\frac{\pi }{4}|\frac{2 b}{2 a+b}\right )-b \left (88 a^2 \cos (2 (c+d x))+b \sin (4 (c+d x)) (28 a+3 b \sin (2 (c+d x)))\right )}{120 d \sqrt{4 a+2 b \sin (2 (c+d x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(184*a^3 + 92*a^2*b + 18*a*b^2 + 9*b^3)*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*(c +
d*x)])/(2*a + b)] - 32*a*(4*a^2 - b^2)*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*(c + d*
x)])/(2*a + b)] - b*(88*a^2*Cos[2*(c + d*x)] + b*(28*a + 3*b*Sin[2*(c + d*x)])*Sin[4*(c + d*x)]))/(120*d*Sqrt[
4*a + 2*b*Sin[2*(c + d*x)]])

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Maple [B]  time = 3.159, size = 1138, normalized size = 4.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(240*a^4*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c)
)*b/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))+64*((2*a+b*sin(2*
d*x+2*c))/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*(-(sin(2*d*
x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*a^3*b-24*a^2*((2*a+b*sin(2*d*x+2*c))/(2*a-b))
^(1/2)*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x
+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b^2-16*a*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*(-(sin(2*d*x+2*
c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),
((2*a-b)/(2*a+b))^(1/2))*b^3-9*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*EllipticF(((2*a+b*sin(2*d*x+2*c))/(2*a-b
))^(1/2),((2*a-b)/(2*a+b))^(1/2))*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*
b^4-368*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2)*EllipticE(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+
b))^(1/2))*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*a^4+56*((2*a+b*sin(2*d*
x+2*c))/(2*a-b))^(1/2)*EllipticE(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*(-(sin(2*d*x+
2*c)-1)*b/(2*a+b))^(1/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*a^2*b^2+9*((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2
)*EllipticE(((2*a+b*sin(2*d*x+2*c))/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*(-(sin(2*d*x+2*c)-1)*b/(2*a+b))^(1
/2)*(-(1+sin(2*d*x+2*c))*b/(2*a-b))^(1/2)*b^4+3*b^4*sin(2*d*x+2*c)^4+28*a*b^3*sin(2*d*x+2*c)^3+44*a^2*b^2*sin(
2*d*x+2*c)^2-3*b^4*sin(2*d*x+2*c)^2-28*sin(2*d*x+2*c)*a*b^3-44*a^2*b^2)/b/cos(2*d*x+2*c)/(4*a+2*b*sin(2*d*x+2*
c))^(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 ) \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^(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} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a^{2}\right )} \sqrt{b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*sin(d*x + c) - a^2)*sqrt(b*cos(d*x + c
)*sin(d*x + c) + a), 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((a+b*cos(d*x+c)*sin(d*x+c))**(5/2),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((a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out