3.6.60 \(\int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{3/2}} \, dx\) [560]

Optimal. Leaf size=160 \[ \frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac {2 a \left (a^2+6 b^2\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt {\cos (c+d x)}}+\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}} \]

[Out]

2/3*b*(3*a^2+4*b^2)*(e*cos(d*x+c))^(3/2)/d/e^3+2*a*b*(e*cos(d*x+c))^(3/2)*(a+b*sin(d*x+c))/d/e^3+2*(b+a*sin(d*
x+c))*(a+b*sin(d*x+c))^2/d/e/(e*cos(d*x+c))^(1/2)-2*a*(a^2+6*b^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^2/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 160, 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 = {2770, 2941, 2748, 2721, 2719} \begin {gather*} \frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac {2 a \left (a^2+6 b^2\right ) 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 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(3*a^2 + 4*b^2)*(e*Cos[c + d*x])^(3/2))/(3*d*e^3) - (2*a*(a^2 + 6*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2])/(d*e^2*Sqrt[Cos[c + d*x]]) + (2*a*b*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x]))/(d*e^3) + (2*(b
+ a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(d*e*Sqrt[e*Cos[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 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[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])

Rule 2770

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*C
os[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Dist[1/(g^2*(p +
1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*S
in[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (Integers
Q[2*m, 2*p] || IntegerQ[m])

Rule 2941

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x
] + Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*
d*m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] &&
GtQ[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin {align*} \int \frac {(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{3/2}} \, dx &=\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {2 \int \sqrt {e \cos (c+d x)} (a+b \sin (c+d x)) \left (\frac {a^2}{2}+2 b^2+\frac {5}{2} a b \sin (c+d x)\right ) \, dx}{e^2}\\ &=\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {4 \int \sqrt {e \cos (c+d x)} \left (\frac {5}{4} a \left (a^2+6 b^2\right )+\frac {5}{4} b \left (3 a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx}{5 e^2}\\ &=\frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}+\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (a \left (a^2+6 b^2\right )\right ) \int \sqrt {e \cos (c+d x)} \, dx}{e^2}\\ &=\frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}+\frac {2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (a \left (a^2+6 b^2\right ) \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{e^2 \sqrt {\cos (c+d x)}}\\ &=\frac {2 b \left (3 a^2+4 b^2\right ) (e \cos (c+d x))^{3/2}}{3 d e^3}-\frac {2 a \left (a^2+6 b^2\right ) \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 b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{d e^3}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{d e \sqrt {e \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.44, size = 98, normalized size = 0.61 \begin {gather*} \frac {2 b^3 \cos ^2(c+d x)-6 a \left (a^2+6 b^2\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+6 \left (3 a^2 b+b^3+a \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{3 d e \sqrt {e \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 8.20, size = 248, normalized size = 1.55

method	result	size
default	\(-\frac {2 \left (-4 b^{3} \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{3}+18 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-6 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18 a \,b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 b^{3} \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 a^{2} b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )-4 b^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 e \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/e/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)*(-4*b^3*sin(1/2*d*x+1/2*c)^5+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))*a^3+18*(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))*a*b^2-6*a^3*cos(1/2*d*x+1/2*c)
*sin(1/2*d*x+1/2*c)^2-18*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+4*b^3*sin(1/2*d*x+1/2*c)^3-9*a^2*b*sin(
1/2*d*x+1/2*c)-4*b^3*sin(1/2*d*x+1/2*c))/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+b*sin(d*x+c))^3/(e*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

e^(-3/2)*integrate((b*sin(d*x + c) + a)^3/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 = 156, normalized size = 0.98 \begin {gather*} -\frac {{\left (3 \, \sqrt {2} {\left (i \, a^{3} + 6 i \, a b^{2}\right )} \cos \left (d x + c\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 \, \sqrt {2} {\left (-i \, a^{3} - 6 i \, a b^{2}\right )} \cos \left (d x + c\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 ) - 2 \, {\left (b^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{2} b + 3 \, b^{3} + 3 \, {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}\right )} e^{\left (-\frac {3}{2}\right )}}{3 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*sqrt(2)*(I*a^3 + 6*I*a*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c
) + I*sin(d*x + c))) + 3*sqrt(2)*(-I*a^3 - 6*I*a*b^2)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(
-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(b^3*cos(d*x + c)^2 + 9*a^2*b + 3*b^3 + 3*(a^3 + 3*a*b^2)*sin(d*x +
c))*sqrt(cos(d*x + c)))*e^(-3/2)/(d*cos(d*x + c))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3065 deep

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

[Out]

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

[Out]

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

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