3.1.40 \(\int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{7/2}} \, dx\) [40]

Optimal. Leaf size=131 \[ -\frac {2 b}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}-\frac {6 a \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {6 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2748, 2716, 2721, 2719} \begin {gather*} -\frac {6 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}}-\frac {6 a \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 a \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 b}{5 d e (e \sin (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])/(e*Sin[c + d*x])^(7/2),x]

[Out]

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

Rule 2716

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

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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])

Rubi steps

\begin {align*} \int \frac {a+b \cos (c+d x)}{(e \sin (c+d x))^{7/2}} \, dx &=-\frac {2 b}{5 d e (e \sin (c+d x))^{5/2}}+a \int \frac {1}{(e \sin (c+d x))^{7/2}} \, dx\\ &=-\frac {2 b}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}+\frac {(3 a) \int \frac {1}{(e \sin (c+d x))^{3/2}} \, dx}{5 e^2}\\ &=-\frac {2 b}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}-\frac {6 a \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {(3 a) \int \sqrt {e \sin (c+d x)} \, dx}{5 e^4}\\ &=-\frac {2 b}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}-\frac {6 a \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {\left (3 a \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 e^4 \sqrt {\sin (c+d x)}}\\ &=-\frac {2 b}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 a \cos (c+d x)}{5 d e (e \sin (c+d x))^{5/2}}-\frac {6 a \cos (c+d x)}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {6 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 74, normalized size = 0.56 \begin {gather*} \frac {-4 b-7 a \cos (c+d x)+3 a \cos (3 (c+d x))+12 a E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right ) \sin ^{\frac {5}{2}}(c+d x)}{10 d e (e \sin (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])/(e*Sin[c + d*x])^(7/2),x]

[Out]

(-4*b - 7*a*Cos[c + d*x] + 3*a*Cos[3*(c + d*x)] + 12*a*EllipticE[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(5/2))
/(10*d*e*(e*Sin[c + d*x])^(5/2))

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Maple [A]
time = 0.14, size = 187, normalized size = 1.43

method	result	size
default	\(\frac {-\frac {2 b}{5 e \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {a \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \left (\sin ^{5}\left (d x +c \right )\right )-4 \left (\sin ^{3}\left (d x +c \right )\right )-2 \sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))/(e*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

(-2/5*b/e/(e*sin(d*x+c))^(5/2)+1/5*a/e^3*(6*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*Elli
pticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*Ellip
ticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+6*sin(d*x+c)^5-4*sin(d*x+c)^3-2*sin(d*x+c))/sin(d*x+c)^3/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))/(e*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

e^(-7/2)*integrate((b*cos(d*x + c) + a)/sin(d*x + c)^(7/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 171, normalized size = 1.31 \begin {gather*} -\frac {3 \, \sqrt {-i} {\left (i \, \sqrt {2} a \cos \left (d x + c\right )^{2} - i \, \sqrt {2} a\right )} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {i} {\left (-i \, \sqrt {2} a \cos \left (d x + c\right )^{2} + i \, \sqrt {2} a\right )} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, a \cos \left (d x + c\right )^{3} - 4 \, a \cos \left (d x + c\right ) - b\right )} \sqrt {\sin \left (d x + c\right )}}{5 \, {\left (d \cos \left (d x + c\right )^{2} e^{\frac {7}{2}} - d e^{\frac {7}{2}}\right )} \sin \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/5*(3*sqrt(-I)*(I*sqrt(2)*a*cos(d*x + c)^2 - I*sqrt(2)*a)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInv
erse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(I)*(-I*sqrt(2)*a*cos(d*x + c)^2 + I*sqrt(2)*a)*sin(d*x + c
)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*a*cos(d*x + c)^3 - 4*
a*cos(d*x + c) - b)*sqrt(sin(d*x + c)))/((d*cos(d*x + c)^2*e^(7/2) - d*e^(7/2))*sin(d*x + c))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))/(e*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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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))/(e*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)*e^(-7/2)/sin(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\cos \left (c+d\,x\right )}{{\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))/(e*sin(c + d*x))^(7/2),x)

[Out]

int((a + b*cos(c + d*x))/(e*sin(c + d*x))^(7/2), x)

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