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(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))^(1/2)-2*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d
*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2)/d/((a+b*sin(
d*x+c))/(a+b))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.30, 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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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \sin \left (d x + c\right ) + a}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [B]  time = 0.25, size = 443, normalized size = 3.99 \[ \frac {2 a^{2} \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )-2 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticF \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}-2 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2}+2 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, \EllipticE \left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2}-2 b^{2} \left (\sin ^{2}\left (d x +c \right )\right )+2 b^{2}}{b \left (a^{2}-b^{2}\right ) \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d} \]

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-b^2*sin(d*x
+c)^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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(c + d*x))^(3/2),x)

[Out]

int(1/(a + b*sin(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

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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