3.565 \(\int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=258 \[ -\frac {10 a b \left (115 a^2+94 b^2\right ) (e \cos (c+d x))^{7/2}}{3003 d e}-\frac {2 b \left (73 a^2+22 b^2\right ) (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{429 d e}+\frac {2 e^2 \left (39 a^4+52 a^2 b^2+4 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{65 d \sqrt {\cos (c+d x)}}+\frac {2 e \left (39 a^4+52 a^2 b^2+4 b^4\right ) \sin (c+d x) (e \cos (c+d x))^{3/2}}{195 d}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e} \]

[Out]

-10/3003*a*b*(115*a^2+94*b^2)*(e*cos(d*x+c))^(7/2)/d/e+2/195*(39*a^4+52*a^2*b^2+4*b^4)*e*(e*cos(d*x+c))^(3/2)*
sin(d*x+c)/d-2/429*b*(73*a^2+22*b^2)*(e*cos(d*x+c))^(7/2)*(a+b*sin(d*x+c))/d/e-38/143*a*b*(e*cos(d*x+c))^(7/2)
*(a+b*sin(d*x+c))^2/d/e-2/13*b*(e*cos(d*x+c))^(7/2)*(a+b*sin(d*x+c))^3/d/e+2/65*(39*a^4+52*a^2*b^2+4*b^4)*e^2*
(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/d/c
os(d*x+c)^(1/2)

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Rubi [A]  time = 0.51, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 e^2 \left (52 a^2 b^2+39 a^4+4 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{65 d \sqrt {\cos (c+d x)}}-\frac {10 a b \left (115 a^2+94 b^2\right ) (e \cos (c+d x))^{7/2}}{3003 d e}-\frac {2 b \left (73 a^2+22 b^2\right ) (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{429 d e}+\frac {2 e \left (52 a^2 b^2+39 a^4+4 b^4\right ) \sin (c+d x) (e \cos (c+d x))^{3/2}}{195 d}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])^4,x]

[Out]

(-10*a*b*(115*a^2 + 94*b^2)*(e*Cos[c + d*x])^(7/2))/(3003*d*e) + (2*(39*a^4 + 52*a^2*b^2 + 4*b^4)*e^2*Sqrt[e*C
os[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(65*d*Sqrt[Cos[c + d*x]]) + (2*(39*a^4 + 52*a^2*b^2 + 4*b^4)*e*(e*Cos[
c + d*x])^(3/2)*Sin[c + d*x])/(195*d) - (2*b*(73*a^2 + 22*b^2)*(e*Cos[c + d*x])^(7/2)*(a + b*Sin[c + d*x]))/(4
29*d*e) - (38*a*b*(e*Cos[c + d*x])^(7/2)*(a + b*Sin[c + d*x])^2)/(143*d*e) - (2*b*(e*Cos[c + d*x])^(7/2)*(a +
b*Sin[c + d*x])^3)/(13*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 (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^4 \, dx &=-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}+\frac {2}{13} \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x))^2 \left (\frac {13 a^2}{2}+3 b^2+\frac {19}{2} a b \sin (c+d x)\right ) \, dx\\ &=-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}+\frac {4}{143} \int (e \cos (c+d x))^{5/2} (a+b \sin (c+d x)) \left (\frac {1}{4} a \left (143 a^2+142 b^2\right )+\frac {3}{4} b \left (73 a^2+22 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac {2 b \left (73 a^2+22 b^2\right ) (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{429 d e}-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}+\frac {8 \int (e \cos (c+d x))^{5/2} \left (\frac {33}{8} \left (39 a^4+52 a^2 b^2+4 b^4\right )+\frac {15}{8} a b \left (115 a^2+94 b^2\right ) \sin (c+d x)\right ) \, dx}{1287}\\ &=-\frac {10 a b \left (115 a^2+94 b^2\right ) (e \cos (c+d x))^{7/2}}{3003 d e}-\frac {2 b \left (73 a^2+22 b^2\right ) (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{429 d e}-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}+\frac {1}{39} \left (39 a^4+52 a^2 b^2+4 b^4\right ) \int (e \cos (c+d x))^{5/2} \, dx\\ &=-\frac {10 a b \left (115 a^2+94 b^2\right ) (e \cos (c+d x))^{7/2}}{3003 d e}+\frac {2 \left (39 a^4+52 a^2 b^2+4 b^4\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{195 d}-\frac {2 b \left (73 a^2+22 b^2\right ) (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{429 d e}-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}+\frac {1}{65} \left (\left (39 a^4+52 a^2 b^2+4 b^4\right ) e^2\right ) \int \sqrt {e \cos (c+d x)} \, dx\\ &=-\frac {10 a b \left (115 a^2+94 b^2\right ) (e \cos (c+d x))^{7/2}}{3003 d e}+\frac {2 \left (39 a^4+52 a^2 b^2+4 b^4\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{195 d}-\frac {2 b \left (73 a^2+22 b^2\right ) (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{429 d e}-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}+\frac {\left (\left (39 a^4+52 a^2 b^2+4 b^4\right ) e^2 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{65 \sqrt {\cos (c+d x)}}\\ &=-\frac {10 a b \left (115 a^2+94 b^2\right ) (e \cos (c+d x))^{7/2}}{3003 d e}+\frac {2 \left (39 a^4+52 a^2 b^2+4 b^4\right ) e^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{65 d \sqrt {\cos (c+d x)}}+\frac {2 \left (39 a^4+52 a^2 b^2+4 b^4\right ) e (e \cos (c+d x))^{3/2} \sin (c+d x)}{195 d}-\frac {2 b \left (73 a^2+22 b^2\right ) (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))}{429 d e}-\frac {38 a b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^2}{143 d e}-\frac {2 b (e \cos (c+d x))^{7/2} (a+b \sin (c+d x))^3}{13 d e}\\ \end {align*}

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Mathematica [A]  time = 2.15, size = 209, normalized size = 0.81 \[ \frac {(e \cos (c+d x))^{5/2} \left (2 \left (39 a^4+52 a^2 b^2+4 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+65 \sqrt {\cos (c+d x)} \left (-\frac {1}{78} b^2 \left (13 a^2+b^2\right ) \sin (4 (c+d x))-\frac {1}{77} a b \left (66 a^2+31 b^2\right ) \cos (c+d x)-\frac {1}{154} a b \left (44 a^2+9 b^2\right ) \cos (3 (c+d x))+\frac {\left (624 a^4-208 a^2 b^2-61 b^4\right ) \sin (2 (c+d x))}{3120}+\frac {1}{22} a b^3 \cos (5 (c+d x))+\frac {1}{208} b^4 \sin (6 (c+d x))\right )\right )}{65 d \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/2)*(a + b*Sin[c + d*x])^4,x]

[Out]

((e*Cos[c + d*x])^(5/2)*(2*(39*a^4 + 52*a^2*b^2 + 4*b^4)*EllipticE[(c + d*x)/2, 2] + 65*Sqrt[Cos[c + d*x]]*(-1
/77*(a*b*(66*a^2 + 31*b^2)*Cos[c + d*x]) - (a*b*(44*a^2 + 9*b^2)*Cos[3*(c + d*x)])/154 + (a*b^3*Cos[5*(c + d*x
)])/22 + ((624*a^4 - 208*a^2*b^2 - 61*b^4)*Sin[2*(c + d*x)])/3120 - (b^2*(13*a^2 + b^2)*Sin[4*(c + d*x)])/78 +
(b^4*Sin[6*(c + d*x)])/208)))/(65*d*Cos[c + d*x]^(5/2))

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{4} e^{2} \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} e^{2} \cos \left (d x + c\right )^{4} + {\left (a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} e^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left (a b^{3} e^{2} \cos \left (d x + c\right )^{4} - {\left (a^{3} b + a b^{3}\right )} e^{2} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral((b^4*e^2*cos(d*x + c)^6 - 2*(3*a^2*b^2 + b^4)*e^2*cos(d*x + c)^4 + (a^4 + 6*a^2*b^2 + b^4)*e^2*cos(d*
x + c)^2 - 4*(a*b^3*e^2*cos(d*x + c)^4 - (a^3*b + a*b^3)*e^2*cos(d*x + c)^2)*sin(d*x + c))*sqrt(e*cos(d*x + c)
), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)^4, x)

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maple [B]  time = 3.00, size = 776, normalized size = 3.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^4,x)

[Out]

2/15015/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^3*(-443520*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^12+12012*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^
(1/2))*a^2*b^2+9009*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))*a^4+924*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))*b^4+492800*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10-246400*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*
c)^8+24024*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+48664*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-24024
*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6006*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-924*b^4*cos(1/2*
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+147840*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^14-1048320*a*b^3*sin(1/2*d*x+
1/2*c)^11+349440*a*b^3*sin(1/2*d*x+1/2*c)^13-137280*a^3*b*sin(1/2*d*x+1/2*c)^9+1173120*a*b^3*sin(1/2*d*x+1/2*c
)^9+274560*a^3*b*sin(1/2*d*x+1/2*c)^7-599040*a*b^3*sin(1/2*d*x+1/2*c)^7-205920*a^3*b*sin(1/2*d*x+1/2*c)^5+1216
80*a*b^3*sin(1/2*d*x+1/2*c)^5+68640*a^3*b*sin(1/2*d*x+1/2*c)^3+3120*a*b^3*sin(1/2*d*x+1/2*c)^3-8580*a^3*b*sin(
1/2*d*x+1/2*c)-3120*a*b^3*sin(1/2*d*x+1/2*c)+616*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-320320*a^2*b^2*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+640640*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-448448*a^2*b^2*c
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+128128*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-12012*a^2*b^2*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (b \sin \left (d x + c\right ) + a\right )}^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)*(a+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(5/2)*(b*sin(d*x + c) + a)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))^4,x)

[Out]

int((e*cos(c + d*x))^(5/2)*(a + b*sin(c + d*x))^4, 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))**(5/2)*(a+b*sin(d*x+c))**4,x)

[Out]

Timed out

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