\(\int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx\) [577]

Optimal result

Integrand size = 20, antiderivative size = 143 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {2} b \cos (2 c+2 d x)}{\left (4 a^2-b^2\right ) d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {2 \sqrt {2} E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}{\left (4 a^2-b^2\right ) d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

[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)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2745, 2743, 21, 2734, 2732} \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {2} b \cos (2 c+2 d x)}{d \left (4 a^2-b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {2 \sqrt {2} \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{d \left (4 a^2-b^2\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

[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 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2743

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] && Integ
erQ[2*n]

Rule 2745

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*} \text {integral}& = \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] (verified)

Time = 0.82 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=\frac {2 \left (b \cos (2 (c+d x))+(2 a+b) E\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}\right )}{\left (4 a^2-b^2\right ) d \sqrt {a+\frac {1}{2} b \sin (2 (c+d x))}} \]

[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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(569\) vs. \(2(161)=322\).

Time = 1.20 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.99

method	result	size
default	\(\frac {16 a^{2} \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right )-4 \operatorname {EllipticF}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, b^{2}-16 \operatorname {EllipticE}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, a^{2}+4 \operatorname {EllipticE}\left (\sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}, \sqrt {\frac {2 a -b}{2 a +b}}\right ) \sqrt {\frac {2 a +b \sin \left (2 d x +2 c \right )}{2 a -b}}\, \sqrt {-\frac {\left (\sin \left (2 d x +2 c \right )-1\right ) b}{2 a +b}}\, \sqrt {-\frac {\left (1+\sin \left (2 d x +2 c \right )\right ) b}{2 a -b}}\, b^{2}-4 \sin \left (2 d x +2 c \right )^{2} b^{2}+4 b^{2}}{b \left (4 a^{2}-b^{2}\right ) \cos \left (2 d x +2 c \right ) \sqrt {4 a +2 b \sin \left (2 d x +2 c \right )}\, d}\)	\(570\)

[In]

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

[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))-EllipticF(((2*a+b*si
n(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)*(-(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*EllipticE(((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)*(-(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+EllipticE(((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)*(-(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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 914, normalized size of antiderivative = 6.39 \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=-\frac {{\left (2 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a^{2} b + {\left (-i \, b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - i \, a b^{2}\right )} \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}}\right )} \sqrt {4 i \, b} \sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 i \, a}{b}} E(\arcsin \left (\sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 i \, a}{b}} {\left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )}\right )\,|\,\frac {4 i \, a b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 8 \, a^{2} - b^{2}}{b^{2}}) - {\left (4 i \, a^{3} + 2 \, a^{2} b + 2 \, {\left (2 i \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{2} b + i \, a b^{2} + {\left (2 \, a b^{2} + i \, b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}}\right )} \sqrt {4 i \, b} \sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 i \, a}{b}} F(\arcsin \left (\sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 i \, a}{b}} {\left (\cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )}\right )\,|\,\frac {4 i \, a b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 8 \, a^{2} - b^{2}}{b^{2}}) + {\left ({\left (i \, b^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + i \, a b^{2}\right )} \sqrt {-4 i \, b} \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 2 \, {\left (a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{2} b\right )} \sqrt {-4 i \, b}\right )} \sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 i \, a}{b}} E(\arcsin \left (\sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 i \, a}{b}} {\left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}\right )\,|\,\frac {-4 i \, a b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 8 \, a^{2} - b^{2}}{b^{2}}) + {\left ({\left (2 \, a^{2} b - i \, a b^{2} + {\left (2 \, a b^{2} - i \, b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {-4 i \, b} \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 \, {\left (-2 i \, a^{3} + a^{2} b + {\left (-2 i \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {-4 i \, b}\right )} \sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 i \, a}{b}} F(\arcsin \left (\sqrt {\frac {b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} - 2 i \, a}{b}} {\left (\cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )}\right )\,|\,\frac {-4 i \, a b \sqrt {-\frac {4 \, a^{2} - b^{2}}{b^{2}}} + 8 \, a^{2} - b^{2}}{b^{2}}) - 2 \, {\left (2 \, b^{3} \cos \left (d x + c\right )^{2} - b^{3}\right )} \sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}}{{\left (4 \, a^{2} b^{3} - b^{5}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (4 \, a^{3} b^{2} - a b^{4}\right )} d} \]

[In]

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

[Out]

-((2*a*b^2*cos(d*x + c)*sin(d*x + c) + 2*a^2*b + (-I*b^3*cos(d*x + c)*sin(d*x + c) - I*a*b^2)*sqrt(-(4*a^2 - b
^2)/b^2))*sqrt(4*I*b)*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b)*elliptic_e(arcsin(sqrt((b*sqrt(-(4*a^2 - b^
2)/b^2) + 2*I*a)/b)*(cos(d*x + c) + I*sin(d*x + c))), (4*I*a*b*sqrt(-(4*a^2 - b^2)/b^2) + 8*a^2 - b^2)/b^2) -
(4*I*a^3 + 2*a^2*b + 2*(2*I*a^2*b + a*b^2)*cos(d*x + c)*sin(d*x + c) - (2*a^2*b + I*a*b^2 + (2*a*b^2 + I*b^3)*
cos(d*x + c)*sin(d*x + c))*sqrt(-(4*a^2 - b^2)/b^2))*sqrt(4*I*b)*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b)*
elliptic_f(arcsin(sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) + 2*I*a)/b)*(cos(d*x + c) + I*sin(d*x + c))), (4*I*a*b*sqrt
(-(4*a^2 - b^2)/b^2) + 8*a^2 - b^2)/b^2) + ((I*b^3*cos(d*x + c)*sin(d*x + c) + I*a*b^2)*sqrt(-4*I*b)*sqrt(-(4*
a^2 - b^2)/b^2) + 2*(a*b^2*cos(d*x + c)*sin(d*x + c) + a^2*b)*sqrt(-4*I*b))*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) -
2*I*a)/b)*elliptic_e(arcsin(sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b)*(cos(d*x + c) - I*sin(d*x + c))), (-
4*I*a*b*sqrt(-(4*a^2 - b^2)/b^2) + 8*a^2 - b^2)/b^2) + ((2*a^2*b - I*a*b^2 + (2*a*b^2 - I*b^3)*cos(d*x + c)*si
n(d*x + c))*sqrt(-4*I*b)*sqrt(-(4*a^2 - b^2)/b^2) - 2*(-2*I*a^3 + a^2*b + (-2*I*a^2*b + a*b^2)*cos(d*x + c)*si
n(d*x + c))*sqrt(-4*I*b))*sqrt((b*sqrt(-(4*a^2 - b^2)/b^2) - 2*I*a)/b)*elliptic_f(arcsin(sqrt((b*sqrt(-(4*a^2
- b^2)/b^2) - 2*I*a)/b)*(cos(d*x + c) - I*sin(d*x + c))), (-4*I*a*b*sqrt(-(4*a^2 - b^2)/b^2) + 8*a^2 - b^2)/b^
2) - 2*(2*b^3*cos(d*x + c)^2 - b^3)*sqrt(b*cos(d*x + c)*sin(d*x + c) + a))/((4*a^2*b^3 - b^5)*d*cos(d*x + c)*s
in(d*x + c) + (4*a^3*b^2 - a*b^4)*d)

Sympy [F]

\[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{3/2}} \, dx=\int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[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)