### 3.56 $$\int \frac{1}{(a+b \sin (c+d x))^{5/2}} \, dx$$

Optimal. Leaf size=231 $\frac{8 a b \cos (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{8 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}$

[Out]

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

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Rubi [A]  time = 0.228373, antiderivative size = 231, 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 = {2664, 2754, 2752, 2663, 2661, 2655, 2653} $\frac{8 a b \cos (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}+\frac{2 b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^{3/2}}-\frac{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{8 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Sin[c + d*x])^(-5/2),x]

[Out]

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

Rule 2664

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 + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*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] && LtQ[m, -1] && 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 \frac{1}{(a+b \sin (c+d x))^{5/2}} \, dx &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}-\frac{2 \int \frac{-\frac{3 a}{2}+\frac{1}{2} b \sin (c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{4 \int \frac{\frac{1}{4} \left (3 a^2+b^2\right )+a b \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\frac{(4 a) \int \sqrt{a+b \sin (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2}-\frac{\int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\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 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\sqrt{\frac{a+b \sin (c+d x)}{a+b}} \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{3 \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}\\ &=\frac{2 b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^{3/2}}+\frac{8 a b \cos (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}+\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 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 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 \left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.904899, size = 166, normalized size = 0.72 $\frac{2 \left (b \cos (c+d x) \left (5 a^2+4 a b \sin (c+d x)-b^2\right )+(a-b) (a+b)^2 \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-4 a (a+b)^2 \left (\frac{a+b \sin (c+d x)}{a+b}\right )^{3/2} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )}{3 d (a-b)^2 (a+b)^2 (a+b \sin (c+d x))^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(-4*a*(a + b)^2*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b))^(3/2) + (a -
b)*(a + b)^2*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*((a + b*Sin[c + d*x])/(a + b))^(3/2) + b*Cos[c +
d*x]*(5*a^2 - b^2 + 4*a*b*Sin[c + d*x])))/(3*(a - b)^2*(a + b)^2*d*(a + b*Sin[c + d*x])^(3/2))

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Maple [A]  time = 0.339, size = 497, normalized size = 2.2 \begin{align*}{\frac{1}{d\cos \left ( dx+c \right ) }\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ({\frac{2}{ \left ( 3\,{a}^{2}-3\,{b}^{2} \right ) b}\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ( \sin \left ( dx+c \right ) +{\frac{a}{b}} \right ) ^{-2}}+{\frac{8\,b \left ( \cos \left ( dx+c \right ) \right ) ^{2}a}{3\, \left ({a}^{2}-{b}^{2} \right ) ^{2}}{\frac{1}{\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}}}+2\,{\frac{3\,{a}^{2}+{b}^{2}}{ \left ( 3\,{a}^{4}-6\,{a}^{2}{b}^{2}+3\,{b}^{4} \right ) \sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \left ({\frac{a}{b}}-1 \right ) \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{{\frac{b \left ( 1-\sin \left ( dx+c \right ) \right ) }{a+b}}}\sqrt{{\frac{ \left ( -1-\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) }+{\frac{8\,ab}{3\, \left ({a}^{2}-{b}^{2} \right ) ^{2}} \left ({\frac{a}{b}}-1 \right ) \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{{\frac{b \left ( 1-\sin \left ( dx+c \right ) \right ) }{a+b}}}\sqrt{{\frac{ \left ( -1-\sin \left ( dx+c \right ) \right ) b}{a-b}}} \left ( \left ( -{\frac{a}{b}}-1 \right ){\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) +{\it EllipticF} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) \right ){\frac{1}{\sqrt{- \left ( -b\sin \left ( dx+c \right ) -a \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}}} \right ){\frac{1}{\sqrt{a+b\sin \left ( dx+c \right ) }}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(d*x+c))^(5/2),x)

[Out]

(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*(2/3/(a^2-b^2)/b*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/(sin(d*x+c)+a
/b)^2+8/3*b*cos(d*x+c)^2/(a^2-b^2)^2*a/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)+2*(3*a^2+b^2)/(3*a^4-6*a^2*b^2+
3*b^4)*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/2)/(
-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))+8/3*a*b/(
a^2-b^2)^2*(a/b-1)*((a+b*sin(d*x+c))/(a-b))^(1/2)*(b*(1-sin(d*x+c))/(a+b))^(1/2)*((-1-sin(d*x+c))*b/(a-b))^(1/
2)/(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)*((-a/b-1)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1
/2))+EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))))/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 \frac{1}{{\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(1/(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 (-\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} - 3 \, a b^{2} +{\left (b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \sin{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))**(5/2),x)

[Out]

Integral((a + b*sin(c + d*x))**(-5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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