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*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)])

________________________________________________________________________________________

Rubi [A]  time = 0.0942034, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 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]

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 \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*}

Mathematica [A]  time = 0.434222, 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])

________________________________________________________________________________________

Maple [B]  time = 3.159, size = 570, normalized size = 4. \begin{align*} 4\,{\frac{1}{ \left ( 4\,{a}^{2}-{b}^{2} \right ) b\cos \left ( 2\,dx+2\,c \right ) \sqrt{4\,a+2\,b\sin \left ( 2\,dx+2\,c \right ) }d} \left ( 4\,{a}^{2}\sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}}\sqrt{-{\frac{ \left ( \sin \left ( 2\,dx+2\,c \right ) -1 \right ) b}{2\,a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( 2\,dx+2\,c \right ) \right ) b}{2\,a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}},\sqrt{{\frac{2\,a-b}{2\,a+b}}} \right ) -\sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}},\sqrt{{\frac{2\,a-b}{2\,a+b}}} \right ) \sqrt{-{\frac{ \left ( \sin \left ( 2\,dx+2\,c \right ) -1 \right ) b}{2\,a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( 2\,dx+2\,c \right ) \right ) b}{2\,a-b}}}{b}^{2}-4\,\sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}},\sqrt{{\frac{2\,a-b}{2\,a+b}}} \right ) \sqrt{-{\frac{ \left ( \sin \left ( 2\,dx+2\,c \right ) -1 \right ) b}{2\,a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( 2\,dx+2\,c \right ) \right ) b}{2\,a-b}}}{a}^{2}+\sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{2\,a+b\sin \left ( 2\,dx+2\,c \right ) }{2\,a-b}}},\sqrt{{\frac{2\,a-b}{2\,a+b}}} \right ) \sqrt{-{\frac{ \left ( \sin \left ( 2\,dx+2\,c \right ) -1 \right ) b}{2\,a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( 2\,dx+2\,c \right ) \right ) b}{2\,a-b}}}{b}^{2}- \left ( \sin \left ( 2\,dx+2\,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*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-sin(2*d*x+2*c)^2*b^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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) \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*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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) \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*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)