### 3.51 $$\int (a+b \sin (c+d x))^{5/2} \, dx$$

Optimal. Leaf size=207 $-\frac{16 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (23 a^2+9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac{16 a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 d}$

[Out]

(-16*a*b*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(15*d) - (2*b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(5*d) +
(2*(23*a^2 + 9*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*d*Sqrt[(a + b*
Sin[c + d*x])/(a + b)]) - (16*a*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d
*x])/(a + b)])/(15*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.259981, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {2656, 2753, 2752, 2663, 2661, 2655, 2653} $-\frac{16 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \left (23 a^2+9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{15 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac{16 a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(-16*a*b*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(15*d) - (2*b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(3/2))/(5*d) +
(2*(23*a^2 + 9*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(15*d*Sqrt[(a + b*
Sin[c + d*x])/(a + b)]) - (16*a*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d
*x])/(a + b)])/(15*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 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 \sin (c+d x))^{5/2} \, dx &=-\frac{2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{2}{5} \int \sqrt{a+b \sin (c+d x)} \left (\frac{1}{2} \left (5 a^2+3 b^2\right )+4 a b \sin (c+d x)\right ) \, dx\\ &=-\frac{16 a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 d}-\frac{2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{4}{15} \int \frac{\frac{1}{4} a \left (15 a^2+17 b^2\right )+\frac{1}{4} b \left (23 a^2+9 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{16 a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 d}-\frac{2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac{1}{15} \left (8 a \left (a^2-b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx+\frac{1}{15} \left (23 a^2+9 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx\\ &=-\frac{16 a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 d}-\frac{2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{\left (\left (23 a^2+9 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{15 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (8 a \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}{15 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{16 a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{15 d}-\frac{2 b \cos (c+d x) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{2 \left (23 a^2+9 b^2\right ) 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)}}{15 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{16 a \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}}}{15 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.939453, size = 185, normalized size = 0.89 $\frac{b \cos (c+d x) \left (-22 a^2-28 a b \sin (c+d x)+3 b^2 \cos (2 (c+d x))-3 b^2\right )+16 a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-2 \left (23 a^2 b+23 a^3+9 a b^2+9 b^3\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{15 d \sqrt{a+b \sin (c+d x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(23*a^3 + 23*a^2*b + 9*a*b^2 + 9*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c +
d*x])/(a + b)] + 16*a*(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)] + b*Cos[c + d*x]*(-22*a^2 - 3*b^2 + 3*b^2*Cos[2*(c + d*x)] - 28*a*b*Sin[c + d*x]))/(15*d*Sqrt[a + b*Sin
[c + d*x]])

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Maple [B]  time = 0.137, size = 890, 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*sin(d*x+c))^(5/2),x)

[Out]

2/15*(15*a^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)*El
lipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+8*((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
^3*b-6*a^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)*Elli
pticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^2-8*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^3-9*((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)*Ellipt
icF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^4-23*((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^4+14*((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)*Elliptic
E(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^2+9*((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))
*b^4+3*b^4*sin(d*x+c)^4+14*a*b^3*sin(d*x+c)^3+11*a^2*b^2*sin(d*x+c)^2-3*b^4*sin(d*x+c)^2-14*a*b^3*sin(d*x+c)-1
1*a^2*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{5}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*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 )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt{b \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*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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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{5}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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