3.573 \(\int (a+a \sin (e+f x))^{3/2} \sqrt{c+d \sin (e+f x)} \, dx\)
Optimal. Leaf size=171 \[ \frac{a^{3/2} (c-7 d) (c+d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{4 d^{3/2} f}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt{a \sin (e+f x)+a}}+\frac{a^2 (c-7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 d f \sqrt{a \sin (e+f x)+a}} \]
[Out]
(a^(3/2)*(c - 7*d)*(c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e +
f*x]])])/(4*d^(3/2)*f) + (a^2*(c - 7*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*d*f*Sqrt[a + a*Sin[e + f*x]]
) - (a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(2*d*f*Sqrt[a + a*Sin[e + f*x]])
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Rubi [A] time = 0.310861, antiderivative size = 171, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used =
{2763, 21, 2770, 2775, 205} \[ \frac{a^{3/2} (c-7 d) (c+d) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{4 d^{3/2} f}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt{a \sin (e+f x)+a}}+\frac{a^2 (c-7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 d f \sqrt{a \sin (e+f x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]],x]
[Out]
(a^(3/2)*(c - 7*d)*(c + d)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e +
f*x]])])/(4*d^(3/2)*f) + (a^2*(c - 7*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*d*f*Sqrt[a + a*Sin[e + f*x]]
) - (a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(2*d*f*Sqrt[a + a*Sin[e + f*x]])
Rule 2763
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
Rule 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])
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]
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]
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^{3/2} \sqrt{c+d \sin (e+f x)} \, dx &=-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt{a+a \sin (e+f x)}}+\frac{\int \frac{\left (-\frac{1}{2} a^2 (c-7 d)-\frac{1}{2} a^2 (c-7 d) \sin (e+f x)\right ) \sqrt{c+d \sin (e+f x)}}{\sqrt{a+a \sin (e+f x)}} \, dx}{2 d}\\ &=-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt{a+a \sin (e+f x)}}-\frac{(a (c-7 d)) \int \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)} \, dx}{4 d}\\ &=\frac{a^2 (c-7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 d f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt{a+a \sin (e+f x)}}-\frac{(a (c-7 d) (c+d)) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx}{8 d}\\ &=\frac{a^2 (c-7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 d f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (c-7 d) (c+d)\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 d f}\\ &=\frac{a^{3/2} (c-7 d) (c+d) \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 d^{3/2} f}+\frac{a^2 (c-7 d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 d f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{2 d f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.619842, size = 247, normalized size = 1.44 \[ \frac{(a (\sin (e+f x)+1))^{3/2} \left (\frac{(c-7 d) (c+d) \left (-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )+\log \left (\sqrt{c+d \sin (e+f x)}+\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )-\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )\right )}{d^{3/2}}-\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \sqrt{c+d \sin (e+f x)} (c+2 d \sin (e+f x)+7 d)}{d}\right )}{8 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]],x]
[Out]
((a*(1 + Sin[e + f*x]))^(3/2)*(((c - 7*d)*(c + d)*(-2*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[
c + d*Sin[e + f*x]]] - ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + Log[Sqr
t[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]))/d^(3/2) - (2*(Cos[(e + f*x)/2] - Sin[(e +
f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*(c + 7*d + 2*d*Sin[e + f*x]))/d))/(8*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]
)^3)
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\sqrt{c+d\sin \left ( fx+e \right ) }\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(1/2),x)
[Out]
int((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(1/2),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \sqrt{d \sin \left (f x + e\right ) + c}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^(3/2)*sqrt(d*sin(f*x + e) + c), x)
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Fricas [B] time = 7.05944, size = 2716, normalized size = 15.88 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
[-1/32*((a*c^2 - 6*a*c*d - 7*a*d^2 + (a*c^2 - 6*a*c*d - 7*a*d^2)*cos(f*x + e) + (a*c^2 - 6*a*c*d - 7*a*d^2)*si
n(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 + 1
28*(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*a*d*cos(f*x + e
)^2 + a*c + 5*a*d + (a*c + 7*a*d)*cos(f*x + e) + (2*a*d*cos(f*x + e) - a*c - 5*a*d)*sin(f*x + e))*sqrt(a*sin(f
*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f), -1/16*((a*c^2 - 6*a*c*d -
7*a*d^2 + (a*c^2 - 6*a*c*d - 7*a*d^2)*cos(f*x + e) + (a*c^2 - 6*a*c*d - 7*a*d^2)*sin(f*x + e))*sqrt(a/d)*arcta
n(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*a*d*cos(f*x + e)^2 + a*c + 5*a*d + (a*c + 7*a*d)*cos(f*x + e) + (2*a*d
*cos(f*x + e) - a*c - 5*a*d)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(d*f*cos(f*x + e
) + d*f*sin(f*x + e) + d*f)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**(3/2)*(c+d*sin(f*x+e))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(1/2),x, algorithm="giac")
[Out]
sage2