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

Optimal. Leaf size=111 \[ \frac {a^{3/2} (c-3 d) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{d^{3/2} f}-\frac {a^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d f \sqrt {a \sin (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2763, 21, 2775, 205} \[ \frac {a^{3/2} (c-3 d) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{d^{3/2} f}-\frac {a^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{d f \sqrt {a \sin (e+f x)+a}} \]

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

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

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Mathematica [B]  time = 0.58, size = 301, normalized size = 2.71 \[ \frac {(a (\sin (e+f x)+1))^{3/2} \left (2 \sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right ) \sqrt {c+d \sin (e+f x)}-2 (c-3 d) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{\sqrt {c+d \sin (e+f x)}}\right )-2 \sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right ) \sqrt {c+d \sin (e+f x)}+c \log \left (\sqrt {c+d \sin (e+f x)}+\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )-3 d \log \left (\sqrt {c+d \sin (e+f x)}+\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )-(c-3 d) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{\sqrt {c+d \sin (e+f x)}}\right )\right )}{2 d^{3/2} f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

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)*(-2*(c - 3*d)*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[
e + f*x]]] - (c - 3*d)*ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] + c*Log[S
qrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]] - 3*d*Log[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi +
2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]] - 2*Sqrt[d]*Cos[(e + f*x)/2]*Sqrt[c + d*Sin[e + f*x]] + 2*Sqrt[d]*Sin[(
e + f*x)/2]*Sqrt[c + d*Sin[e + f*x]]))/(2*d^(3/2)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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fricas [B]  time = 0.93, size = 989, normalized size = 8.91 \[ \text {result too large to display} \]

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]

[-1/8*((a*c - 3*a*d + (a*c - 3*a*d)*cos(f*x + e) + (a*c - 3*a*d)*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*(a*cos(f*x + e) - a*sin(f*x + e) + a)*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), -1/4*((a*c - 3*a*d + (a*c - 3*a*d)*co
s(f*x + e) + (a*c - 3*a*d)*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*(a*cos(f*x + e) -
a*sin(f*x + e) + a)*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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}}}{\sqrt {c +d \sin \left (f x +e \right )}}\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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]

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

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