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

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

[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 = 0.09, 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 = {2669, 2635, 2642, 2641} \[ \frac {10 a e^3 \sin (c+d x) \sqrt {e \cos (c+d x)}}{21 d}+\frac {10 a e^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d \sqrt {e \cos (c+d x)}}+\frac {2 a e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}-\frac {2 b (e \cos (c+d x))^{9/2}}{9 d e} \]

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 2635

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] && Integer
Q[2*n]

Rule 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
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 = 0.90, size = 104, normalized size = 0.84 \[ \frac {e^3 \sqrt {e \cos (c+d x)} \left (\sqrt {\cos (c+d x)} (138 a \sin (c+d x)+18 a \sin (3 (c+d x))-28 b \cos (2 (c+d x))-7 b \cos (4 (c+d x))-21 b)+120 a F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{252 d \sqrt {\cos (c+d x)}} \]

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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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b e^{3} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a e^{3} \cos \left (d x + c\right )^{3}\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

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]

integral((b*e^3*cos(d*x + c)^3*sin(d*x + c) + a*e^3*cos(d*x + c)^3)*sqrt(e*cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}\,{d x} \]

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

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maple [A]  time = 1.40, size = 259, normalized size = 2.09 \[ -\frac {2 e^{4} \left (-224 b \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+144 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+560 b \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-216 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-560 b \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+280 b \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -48 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-70 b \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{63 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

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)

[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 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}\,{d x} \]

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]

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

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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