3.309 \(\int \frac {1}{\sqrt {e \cos (c+d x)} (a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ -\frac {4 \sqrt {e \cos (c+d x)}}{5 a d e \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {e \cos (c+d x)}}{5 d e (a \sin (c+d x)+a)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ -\frac {4 \sqrt {e \cos (c+d x)}}{5 a d e \sqrt {a \sin (c+d x)+a}}-\frac {2 \sqrt {e \cos (c+d x)}}{5 d e (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2671

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*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rule 2672

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*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[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] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 59, normalized size = 0.78 \[ -\frac {2 (2 \sin (c+d x)+3) \sqrt {a (\sin (c+d x)+1)} \sqrt {e \cos (c+d x)}}{5 a^2 d e (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 71, normalized size = 0.93 \[ \frac {2 \, \sqrt {e \cos \left (d x + c\right )} \sqrt {a \sin \left (d x + c\right ) + a} {\left (2 \, \sin \left (d x + c\right ) + 3\right )}}{5 \, {\left (a^{2} d e \cos \left (d x + c\right )^{2} - 2 \, a^{2} d e \sin \left (d x + c\right ) - 2 \, a^{2} d e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^(3/2)/(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)^(3/2)), x)

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maple [A]  time = 0.19, size = 44, normalized size = 0.58 \[ -\frac {2 \left (2 \sin \left (d x +c \right )+3\right ) \cos \left (d x +c \right )}{5 d \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {3}{2}} \sqrt {e \cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(1/2),x)

[Out]

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

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maxima [B]  time = 1.09, size = 211, normalized size = 2.78 \[ -\frac {2 \, {\left (3 \, \sqrt {a} \sqrt {e} + \frac {4 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{5 \, {\left (a^{2} e + \frac {2 \, a^{2} e \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} e \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} \sqrt {-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/5*(3*sqrt(a)*sqrt(e) + 4*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(a)*sqrt(e)*sin(d*x + c)^3
/(cos(d*x + c) + 1)^3 - 3*sqrt(a)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*(sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 1)^2/((a^2*e + 2*a^2*e*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*e*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*sqrt(-sin(d*x + c)/(cos(d*x + c) + 1) + 1))

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mupad [B]  time = 6.38, size = 95, normalized size = 1.25 \[ -\frac {4\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (7\,\cos \left (c+d\,x\right )-\cos \left (3\,c+3\,d\,x\right )+5\,\sin \left (2\,c+2\,d\,x\right )\right )}{5\,a^2\,d\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (15\,\sin \left (c+d\,x\right )-6\,\cos \left (2\,c+2\,d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+10\right )} \]

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))^(3/2)),x)

[Out]

-(4*(a*(sin(c + d*x) + 1))^(1/2)*(7*cos(c + d*x) - cos(3*c + 3*d*x) + 5*sin(2*c + 2*d*x)))/(5*a^2*d*(e*cos(c +
d*x))^(1/2)*(15*sin(c + d*x) - 6*cos(2*c + 2*d*x) - sin(3*c + 3*d*x) + 10))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(d*x+c))**(3/2)/(e*cos(d*x+c))**(1/2),x)

[Out]

Integral(1/((a*(sin(c + d*x) + 1))**(3/2)*sqrt(e*cos(c + d*x))), x)

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