3.565 \(\int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=156 \[ -\frac {3 a (c+d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}-\frac {3 \sqrt {a} (c+d)^2 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{4 \sqrt {d} f} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2770, 2775, 205} \[ -\frac {3 a (c+d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}-\frac {3 \sqrt {a} (c+d)^2 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{4 \sqrt {d} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[a]*(c + d)^2*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]
)])/(4*Sqrt[d]*f) - (3*a*(c + d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*f*Sqrt[a + a*Sin[e + f*x]]) - (a*Co
s[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(2*f*Sqrt[a + a*Sin[e + f*x]])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]

Rule 2775

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2} \, dx &=-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {1}{4} (3 (c+d)) \int \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx\\ &=-\frac {3 a (c+d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {1}{8} \left (3 (c+d)^2\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c+d \sin (e+f x)}} \, dx\\ &=-\frac {3 a (c+d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}-\frac {\left (3 a (c+d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{4 f}\\ &=-\frac {3 \sqrt {a} (c+d)^2 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{4 \sqrt {d} f}-\frac {3 a (c+d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{4 f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 2.05, size = 365, normalized size = 2.34 \[ \frac {\sqrt {a (\sin (e+f x)+1)} \left (-2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x)) (5 c+2 d \sin (e+f x)+3 d)-\frac {3 i (c+d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\log \left (\frac {e^{-i e} \left (2 \sqrt {d} \sqrt {2 c e^{i (e+f x)}-i d \left (-1+e^{2 i (e+f x)}\right )}+2 \sqrt [4]{-1} c-2 (-1)^{3/4} d e^{i (e+f x)}\right )}{\sqrt {d}}\right )-\log \left (\frac {2 f e^{\frac {1}{2} i (e-2 f x)} \left (i \sqrt {d} \sqrt {2 c e^{i (e+f x)}-i d \left (-1+e^{2 i (e+f x)}\right )}+\sqrt [4]{-1} c e^{i (e+f x)}+(-1)^{3/4} d\right )}{\sqrt {d}}\right )\right ) \sqrt {(\cos (e+f x)+i \sin (e+f x)) (c+d \sin (e+f x))}}{\sqrt {d}}\right )}{8 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Sin[e + f*x])]*(((-3*I)*(c + d)^2*(Log[(2*(-1)^(1/4)*c - 2*(-1)^(3/4)*d*E^(I*(e + f*x)) + 2*Sqrt[
d]*Sqrt[2*c*E^(I*(e + f*x)) - I*d*(-1 + E^((2*I)*(e + f*x)))])/(Sqrt[d]*E^(I*e))] - Log[(2*E^((I/2)*(e - 2*f*x
))*((-1)^(3/4)*d + (-1)^(1/4)*c*E^(I*(e + f*x)) + I*Sqrt[d]*Sqrt[2*c*E^(I*(e + f*x)) - I*d*(-1 + E^((2*I)*(e +
f*x)))])*f)/Sqrt[d]])*(Cos[(e + f*x)/2] - I*Sin[(e + f*x)/2])*Sqrt[(Cos[e + f*x] + I*Sin[e + f*x])*(c + d*Sin
[e + f*x])])/Sqrt[d] - 2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(c + d*Sin[e + f*x])*(5*c + 3*d + 2*d*Sin[e + f
*x])))/(8*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]])

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fricas [B]  time = 0.91, size = 1069, normalized size = 6.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(3*(c^2 + 2*c*d + d^2 + (c^2 + 2*c*d + d^2)*cos(f*x + e) + (c^2 + 2*c*d + d^2)*sin(f*x + e))*sqrt(-a/d)*
log((128*a*d^4*cos(f*x + e)^5 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 + 128*(2*a*c*d^3 - a*d^4)*
cos(f*x + e)^4 - 32*(5*a*c^2*d^2 - 14*a*c*d^3 + 13*a*d^4)*cos(f*x + e)^3 - 32*(a*c^3*d - 2*a*c^2*d^2 + 9*a*c*d
^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^4*cos(f*x + e)^4 - c^3*d + 17*c^2*d^2 - 59*c*d^3 + 51*d^4 + 24*(c*d^3 -
d^4)*cos(f*x + e)^3 - 2*(5*c^2*d^2 - 26*c*d^3 + 33*d^4)*cos(f*x + e)^2 - (c^3*d - 7*c^2*d^2 + 31*c*d^3 - 25*d
^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 + c^3*d - 17*c^2*d^2 + 59*c*d^3 - 51*d^4 - 8*(3*c*d^3 - 5*d^4)*cos(f
*x + e)^2 - 2*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(
f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*a*c^3*d + 230*a*c^2*d^2 - 476*a*c*d^3 + 289*a*d^4)*cos(f*x + e) + (128*
a*d^4*cos(f*x + e)^4 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - 256*(a*c*d^3 - a*d^4)*cos(f*x + e
)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a*d^4)*cos(f*x + e)^2 + 32*(a*c^3*d - 7*a*c^2*d^2 + 15*a*c*d^3 - 9*a*d^4
)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)) - 8*(2*d*cos(f*x + e)^2 + (5*c + 3*d)*cos(f*x
+ e) + (2*d*cos(f*x + e) - 5*c - d)*sin(f*x + e) + 5*c + d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)
)/(f*cos(f*x + e) + f*sin(f*x + e) + f), 1/16*(3*(c^2 + 2*c*d + d^2 + (c^2 + 2*c*d + d^2)*cos(f*x + e) + (c^2
+ 2*c*d + d^2)*sin(f*x + e))*sqrt(a/d)*arctan(1/4*(8*d^2*cos(f*x + e)^2 - c^2 + 6*c*d - 9*d^2 - 8*(c*d - d^2)*
sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(a/d)/(2*a*d^2*cos(f*x + e)^3 - (3*a*c*d -
a*d^2)*cos(f*x + e)*sin(f*x + e) - (a*c^2 - a*c*d + 2*a*d^2)*cos(f*x + e))) - 4*(2*d*cos(f*x + e)^2 + (5*c +
3*d)*cos(f*x + e) + (2*d*cos(f*x + e) - 5*c - d)*sin(f*x + e) + 5*c + d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f
*x + e) + c))/(f*cos(f*x + e) + f*sin(f*x + e) + f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a +a \sin \left (f x +e \right )}\, \left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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