3.1.47 \(\int \frac {(a+b \cos (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx\) [47]

Optimal. Leaf size=124 \[ -\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}}+\frac {2 \left (a^2-2 b^2\right ) F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{3 d e^2 \sqrt {e \sin (c+d x)}}-\frac {2 a b \sqrt {e \sin (c+d x)}}{3 d e^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2770, 2748, 2721, 2720} \begin {gather*} \frac {2 \left (a^2-2 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)}}{3 d e^3}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))}{3 d e (e \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.29, size = 76, normalized size = 0.61 \begin {gather*} -\frac {2 \left (2 a b+\left (a^2+b^2\right ) \cos (c+d x)+\left (a^2-2 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)\right )}{3 d e (e \sin (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.13, size = 202, normalized size = 1.63

method	result	size
default	\(\frac {-\frac {4 a b}{3 e \left (e \sin \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-2 b^{2} \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+2 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 e^{2} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-4/3*a*b/e/(e*sin(d*x+c))^(3/2)-1/3/e^2*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(5/2)*Ellipt
icF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2-2*b^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))+2*a^2*cos(d*x+c)^2*sin(d*x+c)+2*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))^2/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate((b*cos(d*x + c) + a)^2/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 = 0.10, size = 168, normalized size = 1.35 \begin {gather*} \frac {\sqrt {-i} {\left (\sqrt {2} {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - \sqrt {2} {\left (a^{2} - 2 \, 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^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - \sqrt {2} {\left (a^{2} - 2 \, 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 (2 \, a b + {\left (a^{2} + 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))^2/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(-I)*(sqrt(2)*(a^2 - 2*b^2)*cos(d*x + c)^2 - sqrt(2)*(a^2 - 2*b^2))*weierstrassPInverse(4, 0, cos(d*x
+ c) + I*sin(d*x + c)) + sqrt(I)*(sqrt(2)*(a^2 - 2*b^2)*cos(d*x + c)^2 - sqrt(2)*(a^2 - 2*b^2))*weierstrassPI
nverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(2*a*b + (a^2 + 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 )^{2}}{\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))**2/(e*sin(d*x+c))**(5/2),x)

[Out]

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

[Out]

integrate((b*cos(d*x + c) + a)^2*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 )}^2}{{\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))^2/(e*sin(c + d*x))^(5/2),x)

[Out]

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

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