3.52 \(\int (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=167 \[ -\frac{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 d}+\frac{8 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

[Out]

(-2*b*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(3*d) + (8*a*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a
+ b*Sin[c + d*x]])/(3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*
b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(3*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.164247, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2656, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \sin (c+d x)}}-\frac{2 b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 d}+\frac{8 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*b*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(3*d) + (8*a*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a
+ b*Sin[c + d*x]])/(3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*
b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(3*d*Sqrt[a + b*Sin[c + d*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 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 \sin (c+d x))^{3/2} \, dx &=-\frac{2 b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 d}+\frac{2}{3} \int \frac{\frac{1}{2} \left (3 a^2+b^2\right )+2 a b \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{2 b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 d}+\frac{1}{3} (4 a) \int \sqrt{a+b \sin (c+d x)} \, dx+\frac{1}{3} \left (-a^2+b^2\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{2 b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 d}+\frac{\left (4 a \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (\left (-a^2+b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{3 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{2 b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{3 d}+\frac{8 a E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 \left (a^2-b^2\right ) F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{3 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.78617, size = 142, normalized size = 0.85 \[ \frac{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-2 b \cos (c+d x) (a+b \sin (c+d x))-8 a (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*b*Cos[c + d*x]*(a + b*Sin[c + d*x]) - 8*a*(a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a
+ b*Sin[c + d*x])/(a + b)] + 2*(a^2 - b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c +
d*x])/(a + b)])/(3*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 0.126, size = 656, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^(3/2),x)

[Out]

2/3*(3*a^3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*Elli
pticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a
+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b
-3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((
a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^2*a-((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b
))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^3-4*(
(a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*
sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3+4*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(
1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^2+sin(d
*x+c)^3*b^3+sin(d*x+c)^2*a*b^2-b^3*sin(d*x+c)-a*b^2)/b/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c) + a)^(3/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sin{\left (c + d x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))**(3/2),x)

[Out]

Integral((a + b*sin(c + d*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^(3/2), x)