3.3.31 \(\int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{11/2}} \, dx\) [231]

Optimal. Leaf size=169 \[ \frac {2 a^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d e^6 \sqrt {\cos (c+d x)}}+\frac {4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac {2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^4-a^4 \sin (c+d x)\right )} \]

[Out]

4/9*a^7*(e*cos(d*x+c))^(3/2)/d/e^7/(a-a*sin(d*x+c))^3-2/15*a^8*(e*cos(d*x+c))^(3/2)/d/e^7/(a^2-a^2*sin(d*x+c))
^2-2/15*a^8*(e*cos(d*x+c))^(3/2)/d/e^7/(a^4-a^4*sin(d*x+c))+2/15*a^4*(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/e^6/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 169, 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 = {2749, 2759, 2760, 2762, 2721, 2719} \begin {gather*} \frac {4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac {2 a^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{15 d e^6 \sqrt {\cos (c+d x)}}-\frac {2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^4-a^4 \sin (c+d x)\right )}-\frac {2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^2-a^2 \sin (c+d x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(11/2),x]

[Out]

(2*a^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*e^6*Sqrt[Cos[c + d*x]]) + (4*a^7*(e*Cos[c + d*x])
^(3/2))/(9*d*e^7*(a - a*Sin[c + d*x])^3) - (2*a^8*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a^2 - a^2*Sin[c + d*x])^2
) - (2*a^8*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a^4 - a^4*Sin[c + d*x]))

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 2749

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

Rule 2759

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

Rule 2760

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

Rule 2762

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{11/2}} \, dx &=\frac {a^8 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^4} \, dx}{e^8}\\ &=\frac {4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac {a^6 \int \frac {\sqrt {e \cos (c+d x)}}{(a-a \sin (c+d x))^2} \, dx}{3 e^6}\\ &=\frac {4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac {2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac {a^5 \int \frac {\sqrt {e \cos (c+d x)}}{a-a \sin (c+d x)} \, dx}{15 e^6}\\ &=\frac {4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac {2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}+\frac {a^4 \int \sqrt {e \cos (c+d x)} \, dx}{15 e^6}\\ &=\frac {4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac {2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}+\frac {\left (a^4 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 e^6 \sqrt {\cos (c+d x)}}\\ &=\frac {2 a^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d e^6 \sqrt {\cos (c+d x)}}+\frac {4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac {2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac {2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.13, size = 66, normalized size = 0.39 \begin {gather*} \frac {4\ 2^{3/4} a^4 \, _2F_1\left (-\frac {9}{4},-\frac {3}{4};-\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{9/4}}{9 d e (e \cos (c+d x))^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(11/2),x]

[Out]

(4*2^(3/4)*a^4*Hypergeometric2F1[-9/4, -3/4, -5/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(9/4))/(9*d*e*(e*C
os[c + d*x])^(9/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(513\) vs. \(2(177)=354\).
time = 7.48, size = 514, normalized size = 3.04

method	result	size
default	\(\frac {2 \left (48 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-96 \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-96 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+192 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+72 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-272 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-24 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+176 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-144 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \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}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+42 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+144 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{45 \left (16 \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-32 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \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}\, e^{5} d}\)	\(514\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x,method=_RETURNVERBOSE)

[Out]

2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2*c)^2+1)/sin(1/
2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^5*(48*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+
1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^8-96*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c
)-96*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1
/2*d*x+1/2*c)^6+192*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+72*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*
d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-272*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x
+1/2*c)-24*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*sin(1/2*d*x+1/2*c)^2+176*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-144*sin(1/2*d*x+1/2*c)^5+3*(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))+42*sin(1/2*d*x+1/2*c)^2*cos
(1/2*d*x+1/2*c)+144*sin(1/2*d*x+1/2*c)^3+4*sin(1/2*d*x+1/2*c))*a^4/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+a*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

e^(-11/2)*integrate((a*sin(d*x + c) + a)^4/cos(d*x + c)^(11/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 421, normalized size = 2.49 \begin {gather*} \frac {3 \, {\left (i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} + 3 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) - 4 i \, \sqrt {2} a^{4} + {\left (-i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) + 4 i \, \sqrt {2} a^{4}\right )} \sin \left (d x + c\right )\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 \, {\left (-i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{3} - 3 i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) + 4 i \, \sqrt {2} a^{4} + {\left (i \, \sqrt {2} a^{4} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} a^{4} \cos \left (d x + c\right ) - 4 i \, \sqrt {2} a^{4}\right )} \sin \left (d x + c\right )\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 (3 \, a^{4} \cos \left (d x + c\right )^{3} - 6 \, a^{4} \cos \left (d x + c\right )^{2} + a^{4} \cos \left (d x + c\right ) + 10 \, a^{4} + {\left (3 \, a^{4} \cos \left (d x + c\right )^{2} + 9 \, a^{4} \cos \left (d x + c\right ) + 10 \, a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{45 \, {\left (d \cos \left (d x + c\right )^{3} e^{\frac {11}{2}} + 3 \, d \cos \left (d x + c\right )^{2} e^{\frac {11}{2}} - 2 \, d \cos \left (d x + c\right ) e^{\frac {11}{2}} - 4 \, d e^{\frac {11}{2}} - {\left (d \cos \left (d x + c\right )^{2} e^{\frac {11}{2}} - 2 \, d \cos \left (d x + c\right ) e^{\frac {11}{2}} - 4 \, d e^{\frac {11}{2}}\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x, algorithm="fricas")

[Out]

1/45*(3*(I*sqrt(2)*a^4*cos(d*x + c)^3 + 3*I*sqrt(2)*a^4*cos(d*x + c)^2 - 2*I*sqrt(2)*a^4*cos(d*x + c) - 4*I*sq
rt(2)*a^4 + (-I*sqrt(2)*a^4*cos(d*x + c)^2 + 2*I*sqrt(2)*a^4*cos(d*x + c) + 4*I*sqrt(2)*a^4)*sin(d*x + c))*wei
erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(-I*sqrt(2)*a^4*cos(d*x + c
)^3 - 3*I*sqrt(2)*a^4*cos(d*x + c)^2 + 2*I*sqrt(2)*a^4*cos(d*x + c) + 4*I*sqrt(2)*a^4 + (I*sqrt(2)*a^4*cos(d*x
+ c)^2 - 2*I*sqrt(2)*a^4*cos(d*x + c) - 4*I*sqrt(2)*a^4)*sin(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInv
erse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*a^4*cos(d*x + c)^3 - 6*a^4*cos(d*x + c)^2 + a^4*cos(d*x + c
) + 10*a^4 + (3*a^4*cos(d*x + c)^2 + 9*a^4*cos(d*x + c) + 10*a^4)*sin(d*x + c))*sqrt(cos(d*x + c)))/(d*cos(d*x
+ c)^3*e^(11/2) + 3*d*cos(d*x + c)^2*e^(11/2) - 2*d*cos(d*x + c)*e^(11/2) - 4*d*e^(11/2) - (d*cos(d*x + c)^2*
e^(11/2) - 2*d*cos(d*x + c)*e^(11/2) - 4*d*e^(11/2))*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+a*sin(d*x+c))**4/(e*cos(d*x+c))**(11/2),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+a*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^4*e^(-11/2)/cos(d*x + c)^(11/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^4/(e*cos(c + d*x))^(11/2),x)

[Out]

int((a + a*sin(c + d*x))^4/(e*cos(c + d*x))^(11/2), x)

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