3.6.39 \(\int (e \cos (c+d x))^{7/2} (a+b \sin (c+d x)) \, dx\) [539]

Optimal. Leaf size=124 \[ -\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e}+\frac {10 a e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {10 a e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a e (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 d} \]

[Out]

-2/9*b*(e*cos(d*x+c))^(9/2)/d/e+2/7*a*e*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/d+10/21*a*e^4*(cos(1/2*d*x+1/2*c)^2)^(
1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/(e*cos(d*x+c))^(1/2)+10/21*a*
e^3*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.06, antiderivative size = 124, 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, 2715, 2721, 2720} \begin {gather*} \frac {10 a e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {10 a e^3 \sin (c+d x) \sqrt {e \cos (c+d x)}}{21 d}+\frac {2 a e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}-\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(e*Cos[c + d*x])^(9/2))/(9*d*e) + (10*a*e^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sqrt[e*C
os[c + d*x]]) + (10*a*e^3*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*a*e*(e*Cos[c + d*x])^(5/2)*Sin[c + d*
x])/(7*d)

Rule 2715

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

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])

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{7/2} (a+b \sin (c+d x)) \, dx &=-\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e}+a \int (e \cos (c+d x))^{7/2} \, dx\\ &=-\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e}+\frac {2 a e (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{7} \left (5 a e^2\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e}+\frac {10 a e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a e (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{21} \left (5 a e^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx\\ &=-\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e}+\frac {10 a e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a e (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {\left (5 a e^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 \sqrt {e \cos (c+d x)}}\\ &=-\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e}+\frac {10 a e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {10 a e^3 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 a e (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end {align*}

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Mathematica [A]
time = 0.85, size = 104, normalized size = 0.84 \begin {gather*} \frac {e^3 \sqrt {e \cos (c+d x)} \left (120 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sqrt {\cos (c+d x)} (-21 b-28 b \cos (2 (c+d x))-7 b \cos (4 (c+d x))+138 a \sin (c+d x)+18 a \sin (3 (c+d x)))\right )}{252 d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Sqrt[e*Cos[c + d*x]]*(120*a*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(-21*b - 28*b*Cos[2*(c + d*x)]
- 7*b*Cos[4*(c + d*x)] + 138*a*Sin[c + d*x] + 18*a*Sin[3*(c + d*x)])))/(252*d*Sqrt[Cos[c + d*x]])

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Maple [A]
time = 4.37, size = 259, normalized size = 2.09

method	result	size
default	\(-\frac {2 e^{4} \left (-224 b \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+144 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+560 b \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-216 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-560 b \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+280 b \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -48 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-70 b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{63 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/63/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^4*(-224*b*sin(1/2*d*x+1/2*c)^11+144*a*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+560*b*sin(1/2*d*x+1/2*c)^9-216*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-560*b*
sin(1/2*d*x+1/2*c)^7+168*a*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+280*b*sin(1/2*d*x+1/2*c)^5+15*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-48*a*cos(1/2*d*x+1/2*
c)*sin(1/2*d*x+1/2*c)^2-70*b*sin(1/2*d*x+1/2*c)^3+7*b*sin(1/2*d*x+1/2*c))/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((e*cos(d*x+c))^(7/2)*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

e^(7/2)*integrate((b*sin(d*x + c) + a)*cos(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 = 109, normalized size = 0.88 \begin {gather*} \frac {-15 i \, \sqrt {2} a e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} a e^{\frac {7}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (7 \, b \cos \left (d x + c\right )^{4} e^{\frac {7}{2}} - 3 \, {\left (3 \, a \cos \left (d x + c\right )^{2} e^{\frac {7}{2}} + 5 \, a e^{\frac {7}{2}}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{63 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/63*(-15*I*sqrt(2)*a*e^(7/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*I*sqrt(2)*a*e^(7/
2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(7*b*cos(d*x + c)^4*e^(7/2) - 3*(3*a*cos(d*x
+ c)^2*e^(7/2) + 5*a*e^(7/2))*sin(d*x + c))*sqrt(cos(d*x + c)))/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4846 deep

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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