3.1.56 \(\int \frac {(a+b \cos (c+d x))^3}{(e \sin (c+d x))^{7/2}} \, dx\) [56]

Optimal. Leaf size=192 \[ -\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}+\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {6 a \left (a^2-2 b^2\right ) E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d e^4 \sqrt {\sin (c+d x)}}-\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{5 d e^5} \]

[Out]

-2/5*(b+a*cos(d*x+c))*(a+b*cos(d*x+c))^2/d/e/(e*sin(d*x+c))^(5/2)-2/5*b*(3*a^2-4*b^2)*(e*sin(d*x+c))^(3/2)/d/e
^5+2/5*(a+b*cos(d*x+c))*(a*b-(3*a^2-4*b^2)*cos(d*x+c))/d/e^3/(e*sin(d*x+c))^(1/2)+6/5*a*(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)*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 = 0.18, antiderivative size = 192, 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, 2940, 2748, 2721, 2719} \begin {gather*} -\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{5 d e^5}-\frac {6 a \left (a^2-2 b^2\right ) 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 {2 \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) (a+b \cos (c+d x))}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 (a \cos (c+d x)+b) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3/(e*Sin[c + d*x])^(7/2),x]

[Out]

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

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])

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 2940

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^m*((d + c*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 - 1)*Simp[a*c*(p
+ 2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, 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))^{7/2}} \, dx &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}-\frac {2 \int \frac {(a+b \cos (c+d x)) \left (-\frac {3 a^2}{2}+2 b^2+\frac {1}{2} a b \cos (c+d x)\right )}{(e \sin (c+d x))^{3/2}} \, dx}{5 e^2}\\ &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}+\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{5 d e^3 \sqrt {e \sin (c+d x)}}+\frac {4 \int \left (-\frac {3}{4} a \left (a^2-2 b^2\right )-\frac {3}{4} b \left (3 a^2-4 b^2\right ) \cos (c+d x)\right ) \sqrt {e \sin (c+d x)} \, dx}{5 e^4}\\ &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}+\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{5 d e^5}-\frac {\left (3 a \left (a^2-2 b^2\right )\right ) \int \sqrt {e \sin (c+d x)} \, dx}{5 e^4}\\ &=-\frac {2 (b+a \cos (c+d x)) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}+\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{5 d e^5}-\frac {\left (3 a \left (a^2-2 b^2\right ) \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+a \cos (c+d x)) (a+b \cos (c+d x))^2}{5 d e (e \sin (c+d x))^{5/2}}+\frac {2 (a+b \cos (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \cos (c+d x)\right )}{5 d e^3 \sqrt {e \sin (c+d x)}}-\frac {6 a \left (a^2-2 b^2\right ) 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)}}-\frac {2 b \left (3 a^2-4 b^2\right ) (e \sin (c+d x))^{3/2}}{5 d e^5}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 130, normalized size = 0.68 \begin {gather*} -\frac {12 a^2 b-6 b^3+a \left (7 a^2+6 b^2\right ) \cos (c+d x)+10 b^3 \cos (2 (c+d x))-3 a^3 \cos (3 (c+d x))+6 a b^2 \cos (3 (c+d x))-12 a \left (a^2-2 b^2\right ) 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])^3/(e*Sin[c + d*x])^(7/2),x]

[Out]

-1/10*(12*a^2*b - 6*b^3 + a*(7*a^2 + 6*b^2)*Cos[c + d*x] + 10*b^3*Cos[2*(c + d*x)] - 3*a^3*Cos[3*(c + d*x)] +
6*a*b^2*Cos[3*(c + d*x)] - 12*a*(a^2 - 2*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(5/2))/(d*e*(e*
Sin[c + d*x])^(5/2))

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Maple [A]
time = 0.15, size = 375, normalized size = 1.95

method	result	size
default	\(\frac {-\frac {2 b \left (5 \left (\cos ^{2}\left (d x +c \right )\right ) b^{2}+3 a^{2}-4 b^{2}\right )}{5 e \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {a \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-12 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}+6 a^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-12 b^{2} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )-8 a^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 b^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )}{5 e^{3} \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(375\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3/(e*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

(-2/5*b/e/(e*sin(d*x+c))^(5/2)*(5*cos(d*x+c)^2*b^2+3*a^2-4*b^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)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2-12*(-sin(d*x+c)+1)^(1/2)*(2*sin(
d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2-3*(-sin(d*x+c)+1)^(1/2)*(2*s
in(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*a^2+6*(-sin(d*x+c)+1)^(1/2)*(
2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(7/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2+6*a^2*cos(d*x+c)^4*sin
(d*x+c)-12*b^2*cos(d*x+c)^4*sin(d*x+c)-8*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)
^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))^3/(e*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

e^(-7/2)*integrate((b*cos(d*x + c) + a)^3/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 = 0.11, size = 251, normalized size = 1.31 \begin {gather*} -\frac {3 \, \sqrt {-i} {\left (\sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )}\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 (\sqrt {2} {\left (-i \, a^{3} + 2 i \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, a^{3} - 2 i \, a b^{2}\right )}\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 (5 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{3} - 2 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \, a^{2} b - 4 \, b^{3} + {\left (4 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )\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))^3/(e*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-1/5*(3*sqrt(-I)*(sqrt(2)*(I*a^3 - 2*I*a*b^2)*cos(d*x + c)^2 + sqrt(2)*(-I*a^3 + 2*I*a*b^2))*sin(d*x + c)*weie
rstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(I)*(sqrt(2)*(-I*a^3 + 2*I
*a*b^2)*cos(d*x + c)^2 + sqrt(2)*(I*a^3 - 2*I*a*b^2))*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4
, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(5*b^3*cos(d*x + c)^2 - 3*(a^3 - 2*a*b^2)*cos(d*x + c)^3 + 3*a^2*b -
4*b^3 + (4*a^3 - 3*a*b^2)*cos(d*x + c))*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))**3/(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))^3/(e*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3*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 {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3}{{\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))^3/(e*sin(c + d*x))^(7/2),x)

[Out]

int((a + b*cos(c + d*x))^3/(e*sin(c + d*x))^(7/2), x)

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