3.578 \(\int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=295 \[ \frac {32 \sqrt {2} a b \cos (2 c+2 d x)}{3 d \left (4 a^2-b^2\right )^2 \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {4 \sqrt {2} b \cos (2 c+2 d x)}{3 d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^{3/2}}-\frac {4 \sqrt {2} \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} F\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{3 d \left (4 a^2-b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {32 \sqrt {2} a \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{3 d \left (4 a^2-b^2\right )^2 \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

[Out]

4/3*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))^(3/2)+32/3*a*b*cos(2*d*x+2*c)*2^(1/2)/(4*a^2
-b^2)^2/d/(2*a+b*sin(2*d*x+2*c))^(1/2)-32/3*a*(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)^2/d/((2*a+b*sin(2*d*x+2*
c))/(2*a+b))^(1/2)+4/3*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*EllipticF(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))/(2*a+b))^(1/2)/(4*a^2-b^2)/d/(2*a+b*sin(2*d*x+2*c))^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2666, 2664, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac {32 \sqrt {2} a b \cos (2 c+2 d x)}{3 d \left (4 a^2-b^2\right )^2 \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {4 \sqrt {2} b \cos (2 c+2 d x)}{3 d \left (4 a^2-b^2\right ) (2 a+b \sin (2 c+2 d x))^{3/2}}-\frac {4 \sqrt {2} \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}} F\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{3 d \left (4 a^2-b^2\right ) \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {32 \sqrt {2} a \sqrt {2 a+b \sin (2 c+2 d x)} E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{3 d \left (4 a^2-b^2\right )^2 \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x]*Sin[c + d*x])^(-5/2),x]

[Out]

(4*Sqrt[2]*b*Cos[2*c + 2*d*x])/(3*(4*a^2 - b^2)*d*(2*a + b*Sin[2*c + 2*d*x])^(3/2)) + (32*Sqrt[2]*a*b*Cos[2*c
+ 2*d*x])/(3*(4*a^2 - b^2)^2*d*Sqrt[2*a + b*Sin[2*c + 2*d*x]]) + (32*Sqrt[2]*a*EllipticE[c - Pi/4 + d*x, (2*b)
/(2*a + b)]*Sqrt[2*a + b*Sin[2*c + 2*d*x]])/(3*(4*a^2 - b^2)^2*d*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)]) -
(4*Sqrt[2]*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*Sqrt[(2*a + b*Sin[2*c + 2*d*x])/(2*a + b)])/(3*(4*a^2 -
b^2)*d*Sqrt[2*a + b*Sin[2*c + 2*d*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 2661

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/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]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {1}{(a+b \cos (c+d x) \sin (c+d x))^{5/2}} \, dx &=\int \frac {1}{\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^{5/2}} \, dx\\ &=\frac {4 \sqrt {2} b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^{3/2}}-\frac {8 \int \frac {-\frac {3 a}{2}+\frac {1}{4} b \sin (2 c+2 d x)}{\left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^{3/2}} \, dx}{3 \left (4 a^2-b^2\right )}\\ &=\frac {4 \sqrt {2} b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^{3/2}}+\frac {32 \sqrt {2} a b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right )^2 d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {64 \int \frac {\frac {1}{16} \left (12 a^2+b^2\right )+\frac {1}{2} a b \sin (2 c+2 d x)}{\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}} \, dx}{3 \left (4 a^2-b^2\right )^2}\\ &=\frac {4 \sqrt {2} b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^{3/2}}+\frac {32 \sqrt {2} a b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right )^2 d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {(64 a) \int \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)} \, dx}{3 \left (4 a^2-b^2\right )^2}-\frac {4 \int \frac {1}{\sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}} \, dx}{3 \left (4 a^2-b^2\right )}\\ &=\frac {4 \sqrt {2} b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^{3/2}}+\frac {32 \sqrt {2} a b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right )^2 d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {\left (64 a \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}{3 \left (4 a^2-b^2\right )^2 \sqrt {\frac {a+\frac {1}{2} b \sin (2 c+2 d x)}{a+\frac {b}{2}}}}-\frac {\left (4 \sqrt {\frac {a+\frac {1}{2} b \sin (2 c+2 d x)}{a+\frac {b}{2}}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+\frac {b}{2}}+\frac {b \sin (2 c+2 d x)}{2 \left (a+\frac {b}{2}\right )}}} \, dx}{3 \left (4 a^2-b^2\right ) \sqrt {a+\frac {1}{2} b \sin (2 c+2 d x)}}\\ &=\frac {4 \sqrt {2} b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right ) d (2 a+b \sin (2 c+2 d x))^{3/2}}+\frac {32 \sqrt {2} a b \cos (2 c+2 d x)}{3 \left (4 a^2-b^2\right )^2 d \sqrt {2 a+b \sin (2 c+2 d x)}}+\frac {32 \sqrt {2} a 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)}}{3 \left (4 a^2-b^2\right )^2 d \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}-\frac {4 \sqrt {2} F\left (c-\frac {\pi }{4}+d x|\frac {2 b}{2 a+b}\right ) \sqrt {\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}}}{3 \left (4 a^2-b^2\right ) d \sqrt {2 a+b \sin (2 c+2 d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.52, size = 201, normalized size = 0.68 \[ -\frac {4 \sqrt {2} \left (b \cos (2 (c+d x)) \left (-20 a^2-8 a b \sin (2 (c+d x))+b^2\right )+(2 a-b) (2 a+b)^2 \left (\frac {2 a+b \sin (2 (c+d x))}{2 a+b}\right )^{3/2} F\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )-\frac {8 a (2 a+b \sin (2 (c+d x)))^2 E\left (c+d x-\frac {\pi }{4}|\frac {2 b}{2 a+b}\right )}{\sqrt {\frac {2 a+b \sin (2 (c+d x))}{2 a+b}}}\right )}{3 d \left (b^2-4 a^2\right )^2 (2 a+b \sin (2 (c+d x)))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x]*Sin[c + d*x])^(-5/2),x]

[Out]

(-4*Sqrt[2]*((-8*a*EllipticE[c - Pi/4 + d*x, (2*b)/(2*a + b)]*(2*a + b*Sin[2*(c + d*x)])^2)/Sqrt[(2*a + b*Sin[
2*(c + d*x)])/(2*a + b)] + (2*a - b)*(2*a + b)^2*EllipticF[c - Pi/4 + d*x, (2*b)/(2*a + b)]*((2*a + b*Sin[2*(c
+ d*x)])/(2*a + b))^(3/2) + b*Cos[2*(c + d*x)]*(-20*a^2 + b^2 - 8*a*b*Sin[2*(c + d*x)])))/(3*(-4*a^2 + b^2)^2
*d*(2*a + b*Sin[2*(c + d*x)])^(3/2))

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fricas [F]  time = 1.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a}}{3 \, a b^{2} \cos \left (d x + c\right )^{4} - 3 \, a b^{2} \cos \left (d x + c\right )^{2} - a^{3} + {\left (b^{3} \cos \left (d x + c\right )^{5} - b^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{2} b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(b*cos(d*x + c)*sin(d*x + c) + a)/(3*a*b^2*cos(d*x + c)^4 - 3*a*b^2*cos(d*x + c)^2 - a^3 + (b^3*
cos(d*x + c)^5 - b^3*cos(d*x + c)^3 - 3*a^2*b*cos(d*x + c))*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^(-5/2), x)

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maple [B]  time = 0.52, size = 1554, normalized size = 5.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x)

[Out]

8/3*(8*sin(2*d*x+2*c)*cos(2*d*x+2*c)^2*a*b^3+(b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2)*(-b/(2*a+b)*sin(2*d*
x+2*c)+b/(2*a+b))^(1/2)*(-b/(2*a-b)*sin(2*d*x+2*c)-b/(2*a-b))^(1/2)*b*(24*EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+
2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a^3+4*EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a
-b)/(2*a+b))^(1/2))*a^2*b-6*EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a*
b^2-EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*b^3-32*EllipticE((b/(2*a-b
)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a^3+8*EllipticE((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*
a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a*b^2)*sin(2*d*x+2*c)+(20*a^2*b^2-b^4)*cos(2*d*x+2*c)^2+48*(b/(2*a-b)*sin
(2*d*x+2*c)+2*a/(2*a-b))^(1/2)*EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))
*(-b/(2*a+b)*sin(2*d*x+2*c)+b/(2*a+b))^(1/2)*(-b/(2*a-b)*sin(2*d*x+2*c)-b/(2*a-b))^(1/2)*a^4+8*(b/(2*a-b)*sin(
2*d*x+2*c)+2*a/(2*a-b))^(1/2)*EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*
(-b/(2*a+b)*sin(2*d*x+2*c)+b/(2*a+b))^(1/2)*(-b/(2*a-b)*sin(2*d*x+2*c)-b/(2*a-b))^(1/2)*a^3*b-12*(b/(2*a-b)*si
n(2*d*x+2*c)+2*a/(2*a-b))^(1/2)*EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2)
)*(-b/(2*a+b)*sin(2*d*x+2*c)+b/(2*a+b))^(1/2)*(-b/(2*a-b)*sin(2*d*x+2*c)-b/(2*a-b))^(1/2)*a^2*b^2-2*(b/(2*a-b)
*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2)*EllipticF((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1
/2))*(-b/(2*a+b)*sin(2*d*x+2*c)+b/(2*a+b))^(1/2)*(-b/(2*a-b)*sin(2*d*x+2*c)-b/(2*a-b))^(1/2)*a*b^3-64*(b/(2*a-
b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2)*(-b/(2*a+b)*sin(2*d*x+2*c)+b/(2*a+b))^(1/2)*(-b/(2*a-b)*sin(2*d*x+2*c)-b/
(2*a-b))^(1/2)*EllipticE((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a^4+16*(b/(2*a-
b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2)*(-b/(2*a+b)*sin(2*d*x+2*c)+b/(2*a+b))^(1/2)*(-b/(2*a-b)*sin(2*d*x+2*c)-b/
(2*a-b))^(1/2)*EllipticE((b/(2*a-b)*sin(2*d*x+2*c)+2*a/(2*a-b))^(1/2),((2*a-b)/(2*a+b))^(1/2))*a^2*b^2)/(2*a+b
*sin(2*d*x+2*c))/(4*a^2-b^2)^2/b/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 \[ \int \frac {1}{{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^(-5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*cos(c + d*x)*sin(c + d*x))^(5/2),x)

[Out]

int(1/(a + b*cos(c + d*x)*sin(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*cos(d*x+c)*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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