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

Optimal. Leaf size=143 \[ \frac {2 \sqrt {2} b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {2 \sqrt {2} \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2666, 2664, 21, 2655, 2653} \[ \frac {2 \sqrt {2} b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {2 \sqrt {2} \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx &=\int \frac {1}{\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^{3/2}} \, dx\\ &=\frac {2 \sqrt {2} b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d \sqrt {2 a+b \sin (2 c+2 d x)}}-\frac {8 \int \frac {-\frac {a}{2}-\frac {1}{4} b \sin (2 c+2 d x)}{\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}} \, dx}{4 a^2-b^2}\\ &=\frac {2 \sqrt {2} b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {4 \int \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)} \, dx}{4 a^2-b^2}\\ &=\frac {2 \sqrt {2} b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {\left (4 \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}\right ) \int \sqrt {\frac {a}{a+\frac {b}{2}}+\frac {b \sin (2 c+2 d x)}{2 \left (a+\frac {b}{2}\right )}} \, dx}{\left (4 a^2-b^2\right ) \sqrt {\frac {a+\frac {1}{2} b \sin (2 c+2 d x)}{a+\frac {b}{2}}}}\\ &=\frac {2 \sqrt {2} b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {2 \sqrt {2} E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}{\left (4 a^2-b^2\right ) d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 101, normalized size = 0.71 \[ \frac {2 \left ((2 a+b) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )+b \cos (2 (c+d x))\right )}{d \left (4 a^2-b^2\right ) \sqrt {a+\frac {1}{2} b \sin (2 (c+d x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}}{b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) - a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(b*cos(d*x + c)*sin(d*x + c) + a)/(b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*a*b*cos(d*x + c)*
sin(d*x + c) - a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.50, size = 570, normalized size = 3.99 \[ \frac {16 a^{2} \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, \EllipticF \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right )-4 \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \EllipticF \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, b^{2}-16 \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \EllipticE \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, a^{2}+4 \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \EllipticE \left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, b^{2}-4 b^{2} \left (\sin ^{2}\left (2 d x +2 c \right )\right )+4 b^{2}}{b \left (4 a^{2}-b^{2}\right ) \cos \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sin \left (2 d x +2 c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))**(3/2),x)

[Out]

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

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