3.6.43 \(\int \frac {(c+d \sin (e+f x))^3}{\sqrt {a+a \sin (e+f x)}} \, dx\) [543]

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

[Out]

-(c-d)^3*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))*2^(1/2)/f/a^(1/2)-4/15*d*(21*c^2-12*c*
d+7*d^2)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-2/5*d*cos(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^(1/2)-2/15
*(9*c-d)*d^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/a/f

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Rubi [A]
time = 0.30, antiderivative size = 178, normalized size of antiderivative = 1.00, 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 = {2857, 3047, 3102, 2830, 2728, 212} \begin {gather*} -\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} \end {gather*}

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 212

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

Rule 2728

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 2830

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*S
in[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 2857

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)/Sqrt[a + b*Sin[e + f*x]])*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], 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 3047

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 3102

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]

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 ) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.42, size = 155, normalized size = 0.87 \begin {gather*} -\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((-60-60 i) (-1)^{3/4} (c-d)^3 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )-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+30 c d-29 d^2+3 d^2 \cos (2 (e+f x))-2 (15 c-d) d \sin (e+f x)\right )\right )}{30 f \sqrt {a (1+\sin (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*((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])))/(f*Sqrt[a*(1 + Sin[e + f*x])])

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Maple [A]
time = 3.92, size = 285, normalized size = 1.60

method	result	size
default	\(-\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (15 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c^{3}-45 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c^{2} d +45 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) c \,d^{2}-15 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) d^{3}+6 \left (a -a \sin \left (f x +e \right )\right )^{\frac {5}{2}} d^{3}-30 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a c \,d^{2}-10 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a \,d^{3}+90 c^{2} d \,a^{2} \sqrt {a -a \sin \left (f x +e \right )}+30 d^{3} a^{2} \sqrt {a -a \sin \left (f x +e \right )}\right )}{15 a^{3} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(285\)

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,method=_RETURNVERBOSE)

[Out]

-1/15*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(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)*a
rctanh(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*(a-a*sin(f*x+e))^(5/2)*d^3-30*(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*d^3*a^2*(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (166) = 332\).
time = 0.38, size = 409, normalized size = 2.30 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 [A]
time = 0.54, size = 294, normalized size = 1.65 \begin {gather*} \frac {\frac {15 \, \sqrt {2} {\left (\sqrt {a} c^{3} - 3 \, \sqrt {a} c^{2} d + 3 \, \sqrt {a} c d^{2} - \sqrt {a} d^{3}\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {15 \, \sqrt {2} {\left (\sqrt {a} c^{3} - 3 \, \sqrt {a} c^{2} d + 3 \, \sqrt {a} c d^{2} - \sqrt {a} d^{3}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {4 \, \sqrt {2} {\left (12 \, a^{\frac {9}{2}} d^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 30 \, a^{\frac {9}{2}} c d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 10 \, a^{\frac {9}{2}} d^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 45 \, a^{\frac {9}{2}} c^{2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, a^{\frac {9}{2}} d^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{5} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{30 \, f} \end {gather*}

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]

1/30*(15*sqrt(2)*(sqrt(a)*c^3 - 3*sqrt(a)*c^2*d + 3*sqrt(a)*c*d^2 - sqrt(a)*d^3)*log(sin(-1/4*pi + 1/2*f*x + 1
/2*e) + 1)/(a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 15*sqrt(2)*(sqrt(a)*c^3 - 3*sqrt(a)*c^2*d + 3*sqrt(a)*c*d
^2 - sqrt(a)*d^3)*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 4*sqrt(2)
*(12*a^(9/2)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^5 - 30*a^(9/2)*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 10*a^(
9/2)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 45*a^(9/2)*c^2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 15*a^(9/2)*d^3*s
in(-1/4*pi + 1/2*f*x + 1/2*e))/(a^5*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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