3.50 \(\int (a+b \cos (c+d x))^3 (e \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=202 \[ \frac {2 a e^2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}+\frac {2 b \left (43 a^2+12 b^2\right ) (e \sin (c+d x))^{7/2}}{231 d e}-\frac {2 a e \left (3 a^2+2 b^2\right ) \cos (c+d x) (e \sin (c+d x))^{3/2}}{15 d}+\frac {2 b (e \sin (c+d x))^{7/2} (a+b \cos (c+d x))^2}{11 d e}+\frac {10 a b (e \sin (c+d x))^{7/2} (a+b \cos (c+d x))}{33 d e} \]

[Out]

-2/15*a*(3*a^2+2*b^2)*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/d+2/231*b*(43*a^2+12*b^2)*(e*sin(d*x+c))^(7/2)/d/e+10/
33*a*b*(a+b*cos(d*x+c))*(e*sin(d*x+c))^(7/2)/d/e+2/11*b*(a+b*cos(d*x+c))^2*(e*sin(d*x+c))^(7/2)/d/e-2/5*a*(3*a
^2+2*b^2)*e^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/sin(d*x+c)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2692, 2862, 2669, 2635, 2640, 2639} \[ \frac {2 a e^2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}+\frac {2 b \left (43 a^2+12 b^2\right ) (e \sin (c+d x))^{7/2}}{231 d e}-\frac {2 a e \left (3 a^2+2 b^2\right ) \cos (c+d x) (e \sin (c+d x))^{3/2}}{15 d}+\frac {2 b (e \sin (c+d x))^{7/2} (a+b \cos (c+d x))^2}{11 d e}+\frac {10 a b (e \sin (c+d x))^{7/2} (a+b \cos (c+d x))}{33 d e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(e*Sin[c + d*x])^(5/2),x]

[Out]

(2*a*(3*a^2 + 2*b^2)*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) - (2*
a*(3*a^2 + 2*b^2)*e*Cos[c + d*x]*(e*Sin[c + d*x])^(3/2))/(15*d) + (2*b*(43*a^2 + 12*b^2)*(e*Sin[c + d*x])^(7/2
))/(231*d*e) + (10*a*b*(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(7/2))/(33*d*e) + (2*b*(a + b*Cos[c + d*x])^2*(e*
Sin[c + d*x])^(7/2))/(11*d*e)

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 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[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])

Rule 2692

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m
])

Rule 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[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 (a+b \cos (c+d x))^3 (e \sin (c+d x))^{5/2} \, dx &=\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}}{11 d e}+\frac {2}{11} \int (a+b \cos (c+d x)) \left (\frac {11 a^2}{2}+2 b^2+\frac {15}{2} a b \cos (c+d x)\right ) (e \sin (c+d x))^{5/2} \, dx\\ &=\frac {10 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{33 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}}{11 d e}+\frac {4}{99} \int \left (\frac {33}{4} a \left (3 a^2+2 b^2\right )+\frac {3}{4} b \left (43 a^2+12 b^2\right ) \cos (c+d x)\right ) (e \sin (c+d x))^{5/2} \, dx\\ &=\frac {2 b \left (43 a^2+12 b^2\right ) (e \sin (c+d x))^{7/2}}{231 d e}+\frac {10 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{33 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}}{11 d e}+\frac {1}{3} \left (a \left (3 a^2+2 b^2\right )\right ) \int (e \sin (c+d x))^{5/2} \, dx\\ &=-\frac {2 a \left (3 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{15 d}+\frac {2 b \left (43 a^2+12 b^2\right ) (e \sin (c+d x))^{7/2}}{231 d e}+\frac {10 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{33 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}}{11 d e}+\frac {1}{5} \left (a \left (3 a^2+2 b^2\right ) e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx\\ &=-\frac {2 a \left (3 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{15 d}+\frac {2 b \left (43 a^2+12 b^2\right ) (e \sin (c+d x))^{7/2}}{231 d e}+\frac {10 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{33 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}}{11 d e}+\frac {\left (a \left (3 a^2+2 b^2\right ) e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 \sqrt {\sin (c+d x)}}\\ &=\frac {2 a \left (3 a^2+2 b^2\right ) e^2 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 \sqrt {\sin (c+d x)}}-\frac {2 a \left (3 a^2+2 b^2\right ) e \cos (c+d x) (e \sin (c+d x))^{3/2}}{15 d}+\frac {2 b \left (43 a^2+12 b^2\right ) (e \sin (c+d x))^{7/2}}{231 d e}+\frac {10 a b (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2}}{33 d e}+\frac {2 b (a+b \cos (c+d x))^2 (e \sin (c+d x))^{7/2}}{11 d e}\\ \end {align*}

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Mathematica [A]  time = 1.39, size = 149, normalized size = 0.74 \[ -\frac {(e \sin (c+d x))^{5/2} \left (1848 \left (3 a^3+2 a b^2\right ) E\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )+\sin ^{\frac {3}{2}}(c+d x) \left (462 a \left (4 a^2+b^2\right ) \cos (c+d x)+5 b \left (12 \left (33 a^2+4 b^2\right ) \cos (2 (c+d x))-396 a^2+154 a b \cos (3 (c+d x))+21 b^2 \cos (4 (c+d x))-69 b^2\right )\right )\right )}{4620 d \sin ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(e*Sin[c + d*x])^(5/2),x]

[Out]

-1/4620*((e*Sin[c + d*x])^(5/2)*(1848*(3*a^3 + 2*a*b^2)*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + (462*a*(4*a^2 +
b^2)*Cos[c + d*x] + 5*b*(-396*a^2 - 69*b^2 + 12*(33*a^2 + 4*b^2)*Cos[2*(c + d*x)] + 154*a*b*Cos[3*(c + d*x)] +
21*b^2*Cos[4*(c + d*x)]))*Sin[c + d*x]^(3/2)))/(d*Sin[c + d*x]^(5/2))

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fricas [F]  time = 1.14, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b^{3} e^{2} \cos \left (d x + c\right )^{5} + 3 \, a b^{2} e^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} b e^{2} \cos \left (d x + c\right ) + {\left (3 \, a^{2} b - b^{3}\right )} e^{2} \cos \left (d x + c\right )^{3} - a^{3} e^{2} + {\left (a^{3} - 3 \, a b^{2}\right )} e^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {e \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-(b^3*e^2*cos(d*x + c)^5 + 3*a*b^2*e^2*cos(d*x + c)^4 - 3*a^2*b*e^2*cos(d*x + c) + (3*a^2*b - b^3)*e^
2*cos(d*x + c)^3 - a^3*e^2 + (a^3 - 3*a*b^2)*e^2*cos(d*x + c)^2)*sqrt(e*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3*(e*sin(d*x + c))^(5/2), x)

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maple [A]  time = 0.43, size = 356, normalized size = 1.76 \[ \frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}} \left (7 \left (\cos ^{2}\left (d x +c \right )\right ) b^{2}+33 a^{2}+4 b^{2}\right )}{77 e}-\frac {e^{3} a \left (10 b^{2} \left (\sin ^{6}\left (d x +c \right )\right )+18 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}+12 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-9 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) a^{2}-6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) b^{2}-6 a^{2} \left (\sin ^{4}\left (d x +c \right )\right )-14 \left (\sin ^{4}\left (d x +c \right )\right ) b^{2}+6 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2}+4 \left (\sin ^{2}\left (d x +c \right )\right ) b^{2}\right )}{15 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(e*sin(d*x+c))^(5/2),x)

[Out]

(2/77/e*b*(e*sin(d*x+c))^(7/2)*(7*cos(d*x+c)^2*b^2+33*a^2+4*b^2)-1/15*e^3*a*(10*b^2*sin(d*x+c)^6+18*(-sin(d*x+
c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*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)^(1/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2-9*(-s
in(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*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)^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*b^2
-6*a^2*sin(d*x+c)^4-14*sin(d*x+c)^4*b^2+6*sin(d*x+c)^2*a^2+4*sin(d*x+c)^2*b^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 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^3*(e*sin(d*x + c))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^(5/2)*(a + b*cos(c + d*x))^3,x)

[Out]

int((e*sin(c + d*x))^(5/2)*(a + b*cos(c + d*x))^3, 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((a+b*cos(d*x+c))**3*(e*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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