3.201 \(\int \frac {a+a \sin (c+d x)}{(e \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=91 \[ -\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d e^2 \sqrt {\cos (c+d x)}}+\frac {2 a}{d e \sqrt {e \cos (c+d x)}}+\frac {2 a \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2669, 2636, 2640, 2639} \[ -\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{d e^2 \sqrt {\cos (c+d x)}}+\frac {2 a}{d e \sqrt {e \cos (c+d x)}}+\frac {2 a \sin (c+d x)}{d e \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])/(e*Cos[c + d*x])^(3/2),x]

[Out]

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

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[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])

Rubi steps

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

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Mathematica [C]  time = 0.98, size = 188, normalized size = 2.07 \[ -\frac {a \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (3 (\cos (d x)-i \sin (d x)) \sqrt {i \sin (2 (c+d x))+\cos (2 (c+d x))+1} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )+(\cos (d x)+i \sin (d x)) \sqrt {i \sin (2 (c+d x))+\cos (2 (c+d x))+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )-6 (\sin (c)+\cos (d x))\right )}{6 d e \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])/(e*Cos[c + d*x])^(3/2),x]

[Out]

-1/6*(a*Csc[c/2]*Sec[c/2]*(-6*(Cos[d*x] + Sin[c]) + 3*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c
] + I*Sin[c])^2)]*(Cos[d*x] - I*Sin[d*x])*Sqrt[1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]] + Hypergeometric2F1[
1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*(Cos[d*x] + I*Sin[d*x])*Sqrt[1 + Cos[2*(c + d*x)] + I*S
in[2*(c + d*x)]]))/(d*e*Sqrt[e*Cos[c + d*x]])

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )} {\left (a \sin \left (d x + c\right ) + a\right )}}{e^{2} \cos \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sin \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)/(e*cos(d*x + c))^(3/2), x)

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maple [A]  time = 0.78, size = 117, normalized size = 1.29 \[ -\frac {2 \left (\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{e \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))/(e*cos(d*x+c))^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a \sin \left (d x + c\right ) + a}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)/(e*cos(d*x + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+a\,\sin \left (c+d\,x\right )}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))/(e*cos(c + d*x))^(3/2),x)

[Out]

int((a + a*sin(c + d*x))/(e*cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \frac {1}{\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sin {\left (c + d x \right )}}{\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))/(e*cos(d*x+c))**(3/2),x)

[Out]

a*(Integral((e*cos(c + d*x))**(-3/2), x) + Integral(sin(c + d*x)/(e*cos(c + d*x))**(3/2), x))

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