3.271 \(\int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=191 \[ -\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^4 \sin (c+d x)+a^4\right )}+\frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^4 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^2 \sin (c+d x)+a^2\right )^2}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a \sin (c+d x)+a)^3}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4} \]

[Out]

2/33*(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)/a^
4/d/(e*cos(d*x+c))^(1/2)-2/15*(e*cos(d*x+c))^(1/2)/d/e/(a+a*sin(d*x+c))^4-14/165*(e*cos(d*x+c))^(1/2)/a/d/e/(a
+a*sin(d*x+c))^3-2/33*(e*cos(d*x+c))^(1/2)/d/e/(a^2+a^2*sin(d*x+c))^2-2/33*(e*cos(d*x+c))^(1/2)/d/e/(a^4+a^4*s
in(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.24, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2681, 2683, 2642, 2641} \[ -\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^4 \sin (c+d x)+a^4\right )}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^2 \sin (c+d x)+a^2\right )^2}+\frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^4 d \sqrt {e \cos (c+d x)}}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a \sin (c+d x)+a)^3}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a \sin (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^4),x]

[Out]

(2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(33*a^4*d*Sqrt[e*Cos[c + d*x]]) - (2*Sqrt[e*Cos[c + d*x]])/(1
5*d*e*(a + a*Sin[c + d*x])^4) - (14*Sqrt[e*Cos[c + d*x]])/(165*a*d*e*(a + a*Sin[c + d*x])^3) - (2*Sqrt[e*Cos[c
+ d*x]])/(33*d*e*(a^2 + a^2*Sin[c + d*x])^2) - (2*Sqrt[e*Cos[c + d*x]])/(33*d*e*(a^4 + a^4*Sin[c + d*x]))

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 2681

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)/(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 2683

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 {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^4} \, dx &=-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a+a \sin (c+d x))^4}+\frac {7 \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^3} \, dx}{15 a}\\ &=-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a+a \sin (c+d x))^4}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a+a \sin (c+d x))^3}+\frac {7 \int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^2} \, dx}{33 a^2}\\ &=-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a+a \sin (c+d x))^4}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a+a \sin (c+d x))^3}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^2+a^2 \sin (c+d x)\right )^2}+\frac {\int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))} \, dx}{11 a^3}\\ &=-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a+a \sin (c+d x))^4}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a+a \sin (c+d x))^3}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^4+a^4 \sin (c+d x)\right )}+\frac {\int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{33 a^4}\\ &=-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a+a \sin (c+d x))^4}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a+a \sin (c+d x))^3}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^4+a^4 \sin (c+d x)\right )}+\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{33 a^4 \sqrt {e \cos (c+d x)}}\\ &=\frac {2 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{33 a^4 d \sqrt {e \cos (c+d x)}}-\frac {2 \sqrt {e \cos (c+d x)}}{15 d e (a+a \sin (c+d x))^4}-\frac {14 \sqrt {e \cos (c+d x)}}{165 a d e (a+a \sin (c+d x))^3}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {2 \sqrt {e \cos (c+d x)}}{33 d e \left (a^4+a^4 \sin (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 66, normalized size = 0.35 \[ -\frac {\sqrt {e \cos (c+d x)} \, _2F_1\left (\frac {1}{4},\frac {19}{4};\frac {5}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{4\ 2^{3/4} a^4 d e \sqrt [4]{\sin (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^4),x]

[Out]

-1/4*(Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[1/4, 19/4, 5/4, (1 - Sin[c + d*x])/2])/(2^(3/4)*a^4*d*e*(1 + Sin[
c + d*x])^(1/4))

________________________________________________________________________________________

fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )}}{a^{4} e \cos \left (d x + c\right )^{5} - 8 \, a^{4} e \cos \left (d x + c\right )^{3} + 8 \, a^{4} e \cos \left (d x + c\right ) - 4 \, {\left (a^{4} e \cos \left (d x + c\right )^{3} - 2 \, a^{4} e \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))/(a^4*e*cos(d*x + c)^5 - 8*a^4*e*cos(d*x + c)^3 + 8*a^4*e*cos(d*x + c) - 4*(a^4*e
*cos(d*x + c)^3 - 2*a^4*e*cos(d*x + c))*sin(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(e*cos(d*x + c))*(a*sin(d*x + c) + a)^4), x)

________________________________________________________________________________________

maple [B]  time = 5.29, size = 762, normalized size = 3.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x)

[Out]

-2/165/(128*sin(1/2*d*x+1/2*c)^14-448*sin(1/2*d*x+1/2*c)^12+672*sin(1/2*d*x+1/2*c)^10-560*sin(1/2*d*x+1/2*c)^8
+280*sin(1/2*d*x+1/2*c)^6-84*sin(1/2*d*x+1/2*c)^4+14*sin(1/2*d*x+1/2*c)^2-1)/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/
2*d*x+1/2*c)^2*e+e)^(1/2)*(640*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)^14-2240*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)^12+640*sin(1/2*d*x+1/2*c)^14*cos(1/2*d*x+1/2*c)+3360*
EllipticF(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)^10-1920*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^12-2800*EllipticF(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)^8+2496*sin(1/2*d*x+1/2*c)^10*cos(1/2*d
*x+1/2*c)+1400*(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)^(
1/2)*sin(1/2*d*x+1/2*c)^6-1792*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-420*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+616*sin(1/2*d*x+1/2*c)^6
*cos(1/2*d*x+1/2*c)+70*EllipticF(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)^2-40*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+240*sin(1/2*d*x+1/2*c)^5-5*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)+160*sin(1/2*d*x+
1/2*c)^2*cos(1/2*d*x+1/2*c)-240*sin(1/2*d*x+1/2*c)^3-28*sin(1/2*d*x+1/2*c))/d

________________________________________________________________________________________

maxima [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(1/(a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {e\,\cos \left (c+d\,x\right )}\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*cos(c + d*x))^(1/2)*(a + a*sin(c + d*x))^4),x)

[Out]

int(1/((e*cos(c + d*x))^(1/2)*(a + a*sin(c + d*x))^4), x)

________________________________________________________________________________________

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(1/(a+a*sin(d*x+c))**4/(e*cos(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________