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

Optimal. Leaf size=178 \[ -\frac{4 d \left (21 c^2-12 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 d^2 (9 c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 a f}-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{2} (c-d)^3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f} \]

[Out]

-((Sqrt[2]*(c - d)^3*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f)) - (4*d*(
21*c^2 - 12*c*d + 7*d^2)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (2*(9*c - d)*d^2*Cos[e + f*x]*Sqrt[a
+ a*Sin[e + f*x]])/(15*a*f) - (2*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*Sqrt[a + a*Sin[e + f*x]])

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Rubi [A]  time = 0.439525, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2778, 2968, 3023, 2751, 2649, 206} \[ -\frac{4 d \left (21 c^2-12 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 d^2 (9 c-d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 a f}-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a \sin (e+f x)+a}}-\frac{\sqrt{2} (c-d)^3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*(c - d)^3*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f)) - (4*d*(
21*c^2 - 12*c*d + 7*d^2)*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (2*(9*c - d)*d^2*Cos[e + f*x]*Sqrt[a
+ a*Sin[e + f*x]])/(15*a*f) - (2*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*Sqrt[a + a*Sin[e + f*x]])

Rule 2778

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-2*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/(b*(2*n
- 1)), Int[((c + d*Sin[e + f*x])^(n - 2)*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n -
3))*Sin[e + f*x], x])/Sqrt[a + b*Sin[e + f*x]], 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, 1] && IntegerQ[2*n]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2751

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

Rule 2649

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^3}{\sqrt{a+a \sin (e+f x)}} \, dx &=-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a+a \sin (e+f x)}}-\frac{\int \frac{(c+d \sin (e+f x)) \left (-a \left (5 c^2-c d+4 d^2\right )-a (9 c-d) d \sin (e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx}{5 a}\\ &=-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a+a \sin (e+f x)}}-\frac{\int \frac{-a c \left (5 c^2-c d+4 d^2\right )+\left (-a c (9 c-d) d-a d \left (5 c^2-c d+4 d^2\right )\right ) \sin (e+f x)-a (9 c-d) d^2 \sin ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{5 a}\\ &=-\frac{2 (9 c-d) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 a f}-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a+a \sin (e+f x)}}-\frac{2 \int \frac{-\frac{1}{2} a^2 \left (15 c^3-3 c^2 d+21 c d^2-d^3\right )-a^2 d \left (21 c^2-12 c d+7 d^2\right ) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx}{15 a^2}\\ &=-\frac{4 d \left (21 c^2-12 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}-\frac{2 (9 c-d) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 a f}-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a+a \sin (e+f x)}}+(c-d)^3 \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{4 d \left (21 c^2-12 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}-\frac{2 (9 c-d) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 a f}-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a+a \sin (e+f x)}}-\frac{\left (2 (c-d)^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{2} (c-d)^3 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}-\frac{4 d \left (21 c^2-12 c d+7 d^2\right ) \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}-\frac{2 (9 c-d) d^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 a f}-\frac{2 d \cos (e+f x) (c+d \sin (e+f x))^2}{5 f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.59945, size = 155, normalized size = 0.87 \[ -\frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-2 d \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-90 c^2-2 d (15 c-d) \sin (e+f x)+30 c d+3 d^2 \cos (2 (e+f x))-29 d^2\right )+(-60-60 i) (-1)^{3/4} (c-d)^3 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{30 f \sqrt{a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((-60 - 60*I)*(-1)^(3/4)*(c - d)^3*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1
+ Tan[(e + f*x)/4])] - 2*d*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-90*c^2 + 30*c*d - 29*d^2 + 3*d^2*Cos[2*(e +
f*x)] - 2*(15*c - d)*d*Sin[e + f*x])))/(30*f*Sqrt[a*(1 + Sin[e + f*x])])

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Maple [A]  time = 0.859, size = 285, normalized size = 1.6 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{15\,{a}^{3}\cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( 15\,{a}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){c}^{3}-45\,{a}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){c}^{2}d+45\,{a}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ) c{d}^{2}-15\,{a}^{5/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){d}^{3}+6\,{d}^{3} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{5/2}-30\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}ac{d}^{2}-10\, \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}a{d}^{3}+90\,{c}^{2}d{a}^{2}\sqrt{a-a\sin \left ( fx+e \right ) }+30\,{a}^{2}{d}^{3}\sqrt{a-a\sin \left ( fx+e \right ) } \right ){\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(15*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)
/a^(1/2))*c^3-45*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c^2*d+45*a^(5/2)*2^(1/2)*
arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*c*d^2-15*a^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2
)*2^(1/2)/a^(1/2))*d^3+6*d^3*(a-a*sin(f*x+e))^(5/2)-30*(a-a*sin(f*x+e))^(3/2)*a*c*d^2-10*(a-a*sin(f*x+e))^(3/2
)*a*d^3+90*c^2*d*a^2*(a-a*sin(f*x+e))^(1/2)+30*a^2*d^3*(a-a*sin(f*x+e))^(1/2))/a^3/cos(f*x+e)/(a+a*sin(f*x+e))
^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}{\sqrt{a \sin \left (f x + e\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.72624, size = 965, normalized size = 5.42 \begin{align*} -\frac{\frac{15 \, \sqrt{2}{\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3} +{\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )} \cos \left (f x + e\right ) +{\left (a c^{3} - 3 \, a c^{2} d + 3 \, a c d^{2} - a d^{3}\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac{2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt{a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt{a}} - 4 \,{\left (3 \, d^{3} \cos \left (f x + e\right )^{3} - 45 \, c^{2} d + 30 \, c d^{2} - 17 \, d^{3} -{\left (15 \, c d^{2} - 4 \, d^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (45 \, c^{2} d - 15 \, c d^{2} + 16 \, d^{3}\right )} \cos \left (f x + e\right ) -{\left (3 \, d^{3} \cos \left (f x + e\right )^{2} - 45 \, c^{2} d + 30 \, c d^{2} - 17 \, d^{3} +{\left (15 \, c d^{2} - d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{30 \,{\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*sqrt(2)*(a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*d^3 + (a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*d^3)*cos(f*x + e
) + (a*c^3 - 3*a*c^2*d + 3*a*c*d^2 - a*d^3)*sin(f*x + e))*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x +
e) + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f
*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) - 4*(3*d^3*cos(f*x + e)^3 - 45*c^2*d
+ 30*c*d^2 - 17*d^3 - (15*c*d^2 - 4*d^3)*cos(f*x + e)^2 - (45*c^2*d - 15*c*d^2 + 16*d^3)*cos(f*x + e) - (3*d^3
*cos(f*x + e)^2 - 45*c^2*d + 30*c*d^2 - 17*d^3 + (15*c*d^2 - d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x +
e) + a))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d \sin{\left (e + f x \right )}\right )^{3}}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.96252, size = 1029, normalized size = 5.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(120*sqrt(2)*(c^3 - 3*c^2*d + 3*c*d^2 - d^3)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*t
an(1/2*f*x + 1/2*e)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*f*x + 1/2*e) + 1)) + ((((((45*a^2*c^2*d*
sgn(tan(1/2*f*x + 1/2*e) + 1) - 15*a^2*c*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1) + 13*a^2*d^3*sgn(tan(1/2*f*x + 1/2*
e) + 1))*tan(1/2*f*x + 1/2*e)/a^9 - 15*(3*a^2*c^2*d*sgn(tan(1/2*f*x + 1/2*e) + 1) - 3*a^2*c*d^2*sgn(tan(1/2*f*
x + 1/2*e) + 1) + a^2*d^3*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^9)*tan(1/2*f*x + 1/2*e) + 10*(9*a^2*c^2*d*sgn(tan(1
/2*f*x + 1/2*e) + 1) - 6*a^2*c*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1) + 4*a^2*d^3*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^
9)*tan(1/2*f*x + 1/2*e) - 10*(9*a^2*c^2*d*sgn(tan(1/2*f*x + 1/2*e) + 1) - 6*a^2*c*d^2*sgn(tan(1/2*f*x + 1/2*e)
+ 1) + 4*a^2*d^3*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^9)*tan(1/2*f*x + 1/2*e) + 15*(3*a^2*c^2*d*sgn(tan(1/2*f*x +
1/2*e) + 1) - 3*a^2*c*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1) + a^2*d^3*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^9)*tan(1/2
*f*x + 1/2*e) - (45*a^2*c^2*d*sgn(tan(1/2*f*x + 1/2*e) + 1) - 15*a^2*c*d^2*sgn(tan(1/2*f*x + 1/2*e) + 1) + 13*
a^2*d^3*sgn(tan(1/2*f*x + 1/2*e) + 1))/a^9)/(a*tan(1/2*f*x + 1/2*e)^2 + a)^(5/2) - (120*sqrt(2)*a^10*c^3*arcta
n(sqrt(a)/sqrt(-a)) - 360*sqrt(2)*a^10*c^2*d*arctan(sqrt(a)/sqrt(-a)) + 360*sqrt(2)*a^10*c*d^2*arctan(sqrt(a)/
sqrt(-a)) - 120*sqrt(2)*a^10*d^3*arctan(sqrt(a)/sqrt(-a)) - 45*sqrt(2)*sqrt(-a)*sqrt(a)*c^2*d + 30*sqrt(2)*sqr
t(-a)*sqrt(a)*c*d^2 - 17*sqrt(2)*sqrt(-a)*sqrt(a)*d^3)*sgn(tan(1/2*f*x + 1/2*e) + 1)/(sqrt(-a)*a^10))/f