3.211 \(\int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=127 \[ \frac {2 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d e^4 \sqrt {\cos (c+d x)}}+\frac {2 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^2-a^2 \sin (c+d x)\right )} \]

[Out]

2/5*a^4*(e*cos(d*x+c))^(3/2)/d/e^5/(a-a*sin(d*x+c))^2+2/5*a^4*(e*cos(d*x+c))^(3/2)/d/e^5/(a^2-a^2*sin(d*x+c))-
2/5*a^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/e^4/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2670, 2681, 2683, 2640, 2639} \[ \frac {2 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^2-a^2 \sin (c+d x)\right )}+\frac {2 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac {2 a^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d e^4 \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^2/(e*Cos[c + d*x])^(7/2),x]

[Out]

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

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 2670

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

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Mathematica [C]  time = 0.09, size = 66, normalized size = 0.52 \[ \frac {2\ 2^{3/4} a^2 (\sin (c+d x)+1)^{5/4} \, _2F_1\left (-\frac {5}{4},\frac {1}{4};-\frac {1}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{5 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^2/(e*Cos[c + d*x])^(7/2),x]

[Out]

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

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \sqrt {e \cos \left (d x + c\right )}}{e^{4} \cos \left (d x + c\right )^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^2/(e*cos(d*x + c))^(7/2), x)

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maple [B]  time = 1.38, size = 305, normalized size = 2.40 \[ -\frac {2 \left (4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(7/2),x)

[Out]

-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*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^3*(4*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin
(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-4*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticE(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)+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)-6*sin
(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c))*a^2/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}{\left (e \cos \left (d x + c\right )\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^2/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^2/(e*cos(d*x + c))^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^2/(e*cos(c + d*x))^(7/2),x)

[Out]

int((a + a*sin(c + d*x))^2/(e*cos(c + d*x))^(7/2), 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+a*sin(d*x+c))**2/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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