3.3.28 \(\int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx\) [228]

Optimal. Leaf size=152 \[ -\frac {10 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {e \cos (c+d x)}}-\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^8 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \]

[Out]

4/3*a^7*(e*cos(d*x+c))^(9/2)/d/e^7/(a-a*sin(d*x+c))^3+12*a^8*(e*cos(d*x+c))^(5/2)/d/e^5/(a^4-a^4*sin(d*x+c))-1
0*a^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^2/(e*cos(d*x+c))^(1/2)-10*a^4*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/d/e^3

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Rubi [A]
time = 0.15, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2749, 2759, 2715, 2721, 2720} \begin {gather*} \frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}-\frac {10 a^4 \sin (c+d x) \sqrt {e \cos (c+d x)}}{d e^3}-\frac {10 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {e \cos (c+d x)}}+\frac {12 a^8 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^4-a^4 \sin (c+d x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*e^2*Sqrt[e*Cos[c + d*x]]) - (10*a^4*Sqrt[e*Cos[c + d
*x]]*Sin[c + d*x])/(d*e^3) + (4*a^7*(e*Cos[c + d*x])^(9/2))/(3*d*e^7*(a - a*Sin[c + d*x])^3) + (12*a^8*(e*Cos[
c + d*x])^(5/2))/(d*e^5*(a^4 - a^4*Sin[c + d*x]))

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 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]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx &=\frac {a^8 \int \frac {(e \cos (c+d x))^{11/2}}{(a-a \sin (c+d x))^4} \, dx}{e^8}\\ &=\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}-\frac {\left (3 a^6\right ) \int \frac {(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^2} \, dx}{e^6}\\ &=\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (15 a^4\right ) \int (e \cos (c+d x))^{3/2} \, dx}{e^4}\\ &=-\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{e^2}\\ &=-\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (5 a^4 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{e^2 \sqrt {e \cos (c+d x)}}\\ &=-\frac {10 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {e \cos (c+d x)}}-\frac {10 a^4 \sqrt {e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac {4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac {12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.06, size = 66, normalized size = 0.43 \begin {gather*} \frac {16 \sqrt [4]{2} a^4 \, _2F_1\left (-\frac {9}{4},-\frac {3}{4};\frac {1}{4};\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{3 d e (e \cos (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16*2^(1/4)*a^4*Hypergeometric2F1[-9/4, -3/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(3/4))/(3*d*e*(e*C
os[c + d*x])^(3/2))

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Maple [A]
time = 3.56, size = 263, normalized size = 1.73

method	result	size
default	\(\frac {2 \left (-8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-18 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+48 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{4}}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\)	\(263\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/(2*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^2*(-8*sin(1/2*d*x+1/2*
c)^6*cos(1/2*d*x+1/2*c)+30*(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))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-48*sin(1/2*d*x+1/2*c)^5-15*(si
n(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))-18*sin(1/2*d*
x+1/2*c)^2*cos(1/2*d*x+1/2*c)+48*sin(1/2*d*x+1/2*c)^3-20*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))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 146, normalized size = 0.96 \begin {gather*} -\frac {15 \, {\left (-i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) + i \, \sqrt {2} a^{4}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, {\left (i \, \sqrt {2} a^{4} \sin \left (d x + c\right ) - i \, \sqrt {2} a^{4}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} - 11 \, a^{4} \sin \left (d x + c\right ) + 19 \, a^{4}\right )} \sqrt {\cos \left (d x + c\right )}}{3 \, {\left (d e^{\frac {5}{2}} \sin \left (d x + c\right ) - d e^{\frac {5}{2}}\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))^(5/2),x, algorithm="fricas")

[Out]

-1/3*(15*(-I*sqrt(2)*a^4*sin(d*x + c) + I*sqrt(2)*a^4)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
)) + 15*(I*sqrt(2)*a^4*sin(d*x + c) - I*sqrt(2)*a^4)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))
+ 2*(a^4*cos(d*x + c)^2 - 11*a^4*sin(d*x + c) + 19*a^4)*sqrt(cos(d*x + c)))/(d*e^(5/2)*sin(d*x + c) - d*e^(5/
2))

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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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^4*e^(-5/2)/cos(d*x + c)^(5/2), 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 )}^{5/2}} \,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))^(5/2),x)

[Out]

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

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