### 3.55 $$\int \frac{1}{(a+b \sin (c+d x))^{3/2}} \, dx$$

Optimal. Leaf size=111 $\frac{2 b \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{2 \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}$

[Out]

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

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Rubi [A]  time = 0.0648819, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.286, Rules used = {2664, 21, 2655, 2653} $\frac{2 b \cos (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{2 \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*Sin[c + d*x])^(-3/2),x]

[Out]

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

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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))^{3/2}} \, dx &=\frac{2 b \cos (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{2 \int \frac{-\frac{a}{2}-\frac{1}{2} b \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{2 b \cos (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{\int \sqrt{a+b \sin (c+d x)} \, dx}{a^2-b^2}\\ &=\frac{2 b \cos (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{\sqrt{a+b \sin (c+d x)} \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{\left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}\\ &=\frac{2 b \cos (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}+\frac{2 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)}}{\left (a^2-b^2\right ) d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}\\ \end{align*}

Mathematica [A]  time = 0.289202, size = 87, normalized size = 0.78 $\frac{2 b \cos (c+d x)-2 (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{d (a-b) (a+b) \sqrt{a+b \sin (c+d x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.158, size = 443, normalized size = 4. \begin{align*} 2\,{\frac{1}{ \left ({a}^{2}-{b}^{2} \right ) b\cos \left ( dx+c \right ) \sqrt{a+b\sin \left ( dx+c \right ) }d} \left ({a}^{2}\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{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 ) -\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{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 ){b}^{2}-\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){a}^{2}+\sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( dx+c \right ) -1 \right ) b}{a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) b}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( dx+c \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){b}^{2}- \left ( \sin \left ( dx+c \right ) \right ) ^{2}{b}^{2}+{b}^{2} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(d*x+c))^(3/2),x)

[Out]

2/b*(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)*Ellipt
icF(((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))*b^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)*EllipticE(((a+b*si
n(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*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)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^2-sin(d*x+c)^
2*b^2+b^2)/(a^2-b^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{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 (-\frac{\sqrt{b \sin \left (d x + c\right ) + a}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*sin(d*x + c) + a)/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2), 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{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 \frac{1}{{\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(1/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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