3.1.55 \(\int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{5/2}} \, dx\) [55]

Optimal. Leaf size=169 \[ -\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a \left (a^2-6 b^2\right ) F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 b \left (a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{3 d e^3}-\frac {2 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e^3} \]

[Out]

-2/3*(b+a*cos(d*x+c))*(a+b*cos(d*x+c))^2/d/e/(e*sin(d*x+c))^(3/2)-2/3*a*(a^2-6*b^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)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/d/e^2/(e*sin
(d*x+c))^(1/2)-2/3*b*(a^2+4*b^2)*(e*sin(d*x+c))^(1/2)/d/e^3-2/3*a*b*(a+b*cos(d*x+c))*(e*sin(d*x+c))^(1/2)/d/e^
3

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Rubi [A]
time = 0.16, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2770, 2941, 2748, 2721, 2720} \begin {gather*} -\frac {2 b \left (a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{3 d e^3}+\frac {2 a \left (a^2-6 b^2\right ) \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 a b \sqrt {e \sin (c+d x)} (a+b \cos (c+d x))}{3 d e^3}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(5/2),x]

[Out]

(-2*(b + a*Cos[c + d*x])*(a + b*Cos[c + d*x])^2)/(3*d*e*(e*Sin[c + d*x])^(3/2)) + (2*a*(a^2 - 6*b^2)*EllipticF
[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*e^2*Sqrt[e*Sin[c + d*x]]) - (2*b*(a^2 + 4*b^2)*Sqrt[e*Sin[c +
d*x]])/(3*d*e^3) - (2*a*b*(a + b*Cos[c + d*x])*Sqrt[e*Sin[c + d*x]])/(3*d*e^3)

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2770

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*C
os[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Dist[1/(g^2*(p +
1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*S
in[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (Integers
Q[2*m, 2*p] || IntegerQ[m])

Rule 2941

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

Rubi steps

\begin {align*} \int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{5/2}} \, dx &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 \int \frac {(a+b \cos (c+d x)) \left (-\frac {a^2}{2}+2 b^2+\frac {3}{2} a b \cos (c+d x)\right )}{\sqrt {e \sin (c+d x)}} \, dx}{3 e^2}\\ &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e^3}-\frac {4 \int \frac {-\frac {3}{4} a \left (a^2-6 b^2\right )+\frac {3}{4} b \left (a^2+4 b^2\right ) \cos (c+d x)}{\sqrt {e \sin (c+d x)}} \, dx}{9 e^2}\\ &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 b \left (a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{3 d e^3}-\frac {2 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e^3}+\frac {\left (a \left (a^2-6 b^2\right )\right ) \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{3 e^2}\\ &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac {2 b \left (a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{3 d e^3}-\frac {2 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e^3}+\frac {\left (a \left (a^2-6 b^2\right ) \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{3 e^2 \sqrt {e \sin (c+d x)}}\\ &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 a \left (a^2-6 b^2\right ) F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 b \left (a^2+4 b^2\right ) \sqrt {e \sin (c+d x)}}{3 d e^3}-\frac {2 a b (a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}}{3 d e^3}\\ \end {align*}

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Mathematica [A]
time = 0.92, size = 102, normalized size = 0.60 \begin {gather*} -\frac {6 a^2 b+5 b^3+2 a \left (a^2+3 b^2\right ) \cos (c+d x)-3 b^3 \cos (2 (c+d x))+2 a \left (a^2-6 b^2\right ) F\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {3}{2}}(c+d x)}{3 d e (e \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(5/2),x]

[Out]

-1/3*(6*a^2*b + 5*b^3 + 2*a*(a^2 + 3*b^2)*Cos[c + d*x] - 3*b^3*Cos[2*(c + d*x)] + 2*a*(a^2 - 6*b^2)*EllipticF[
(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(3/2))/(d*e*(e*Sin[c + d*x])^(3/2))

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Maple [A]
time = 0.14, size = 226, normalized size = 1.34

method	result	size
default	\(\frac {-\frac {2 b \left (-3 \left (\cos ^{2}\left (d x +c \right )\right ) b^{2}+3 a^{2}+4 b^{2}\right )}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-6 b^{2} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(226\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-2/3*b/e/(e*sin(d*x+c))^(3/2)*(-3*cos(d*x+c)^2*b^2+3*a^2+4*b^2)-1/3*a/e^2*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c
)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2-6*b^2*(-sin(d*x+c)+1)^(1/2)*(2*si
n(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*a^2*cos(d*x+c)^2*sin(d*x+c)+
6*b^2*cos(d*x+c)^2*sin(d*x+c))/sin(d*x+c)^2/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate((b*cos(d*x + c) + a)^3/sin(d*x + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 196, normalized size = 1.16 \begin {gather*} \frac {\sqrt {-i} {\left (\sqrt {2} {\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - \sqrt {2} {\left (a^{3} - 6 \, a b^{2}\right )}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {i} {\left (\sqrt {2} {\left (a^{3} - 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - \sqrt {2} {\left (a^{3} - 6 \, a b^{2}\right )}\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (3 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - 4 \, b^{3} - {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\sin \left (d x + c\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{2} e^{\frac {5}{2}} - d e^{\frac {5}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(-I)*(sqrt(2)*(a^3 - 6*a*b^2)*cos(d*x + c)^2 - sqrt(2)*(a^3 - 6*a*b^2))*weierstrassPInverse(4, 0, cos
(d*x + c) + I*sin(d*x + c)) + sqrt(I)*(sqrt(2)*(a^3 - 6*a*b^2)*cos(d*x + c)^2 - sqrt(2)*(a^3 - 6*a*b^2))*weier
strassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(3*b^3*cos(d*x + c)^2 - 3*a^2*b - 4*b^3 - (a^3 + 3*a*b
^2)*cos(d*x + c))*sqrt(sin(d*x + c)))/(d*cos(d*x + c)^2*e^(5/2) - d*e^(5/2))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \cos {\left (c + d x \right )}\right )^{3}}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3/(e*sin(d*x+c))**(5/2),x)

[Out]

Integral((a + b*cos(c + d*x))**3/(e*sin(c + d*x))**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3*e^(-5/2)/sin(d*x + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^3/(e*sin(c + d*x))^(5/2),x)

[Out]

int((a + b*cos(c + d*x))^3/(e*sin(c + d*x))^(5/2), x)

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