3.3.9 \(\int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{3/2}} \, dx\) [209]

Optimal. Leaf size=85 \[ -\frac {6 a^2 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {\cos (c+d x)}}+\frac {4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \left (a^2-a^2 \sin (c+d x)\right )} \]

[Out]

4*a^4*(e*cos(d*x+c))^(3/2)/d/e^3/(a^2-a^2*sin(d*x+c))-6*a^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticE(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 = 0.10, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2749, 2759, 2721, 2719} \begin {gather*} \frac {4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {6 a^2 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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^2}{(e \cos (c+d x))^{3/2}} \, dx &=\frac {a^4 \int \frac {(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^2} \, dx}{e^4}\\ &=\frac {4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (3 a^2\right ) \int \sqrt {e \cos (c+d x)} \, dx}{e^2}\\ &=\frac {4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac {\left (3 a^2 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{e^2 \sqrt {\cos (c+d x)}}\\ &=-\frac {6 a^2 \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 {4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \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 = 0.04, size = 64, normalized size = 0.75 \begin {gather*} \frac {4\ 2^{3/4} a^2 \, _2F_1\left (-\frac {3}{4},-\frac {1}{4};\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right ) \sqrt [4]{1+\sin (c+d x)}}{d e \sqrt {e \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 2.44, size = 120, normalized size = 1.41

method	result	size
default	\(-\frac {2 \left (3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{e \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*(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))-4*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 \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))^2/(e*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 184, normalized size = 2.16 \begin {gather*} \frac {3 \, {\left (-i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) + i \, \sqrt {2} a^{2} \sin \left (d x + c\right ) - i \, \sqrt {2} a^{2}\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^{2} \cos \left (d x + c\right ) - i \, \sqrt {2} a^{2} \sin \left (d x + c\right ) + i \, \sqrt {2} a^{2}\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 ) + 4 \, {\left (a^{2} \cos \left (d x + c\right ) + a^{2} \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {\cos \left (d x + c\right )}}{d \cos \left (d x + c\right ) e^{\frac {3}{2}} - d e^{\frac {3}{2}} \sin \left (d x + c\right ) + d e^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(-I*sqrt(2)*a^2*cos(d*x + c) + I*sqrt(2)*a^2*sin(d*x + c) - I*sqrt(2)*a^2)*weierstrassZeta(-4, 0, weierstra
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(I*sqrt(2)*a^2*cos(d*x + c) - I*sqrt(2)*a^2*sin(d*x + c)
+ I*sqrt(2)*a^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 4*(a^2*c
os(d*x + c) + a^2*sin(d*x + c) + a^2)*sqrt(cos(d*x + c)))/(d*cos(d*x + c)*e^(3/2) - d*e^(3/2)*sin(d*x + c) + d
*e^(3/2))

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

[Out]

integrate((a*sin(d*x + c) + a)^2*e^(-3/2)/cos(d*x + c)^(3/2), 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 )}^2}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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