\(\int (a+b \sin (c+d x))^{3/2} \, dx\) [78]

Optimal result

Integrand size = 14, antiderivative size = 167 \[ \int (a+b \sin (c+d x))^{3/2} \, dx=-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}+\frac {8 a E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{3 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 d \sqrt {a+b \sin (c+d x)}} \] Output:

-2/3*b*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/d-8/3*a*EllipticE(cos(1/2*c+1/4*P 
i+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/d/((a+b*sin(d*x 
+c))/(a+b))^(1/2)-2/3*(a^2-b^2)*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/ 
2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/d/(a+b*sin(d*x+c))^(1/2 
)
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.85 \[ \int (a+b \sin (c+d x))^{3/2} \, dx=\frac {-2 b \cos (c+d x) (a+b \sin (c+d x))-8 a (a+b) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+2 \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 d \sqrt {a+b \sin (c+d x)}} \] Input:

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

Output:

(-2*b*Cos[c + d*x]*(a + b*Sin[c + d*x]) - 8*a*(a + b)*EllipticE[(-2*c + Pi 
 - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + 2*(a^2 - 
b^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d 
*x])/(a + b)])/(3*d*Sqrt[a + b*Sin[c + d*x]])
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 3135, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (a+b \sin (c+d x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \sin (c+d x))^{3/2}dx\)

\(\Big \downarrow \) 3135

\(\displaystyle \frac {2}{3} \int \frac {3 a^2+4 b \sin (c+d x) a+b^2}{2 \sqrt {a+b \sin (c+d x)}}dx-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {3 a^2+4 b \sin (c+d x) a+b^2}{\sqrt {a+b \sin (c+d x)}}dx-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {3 a^2+4 b \sin (c+d x) a+b^2}{\sqrt {a+b \sin (c+d x)}}dx-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {1}{3} \left (4 a \int \sqrt {a+b \sin (c+d x)}dx-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (4 a \int \sqrt {a+b \sin (c+d x)}dx-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{3} \left (\frac {4 a \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {4 a \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{\sqrt {a+b \sin (c+d x)}}\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{3} \left (\frac {8 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \sin (c+d x)}}\right )-\frac {2 b \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 d}\)

Input:

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

Output:

(-2*b*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(3*d) + ((8*a*EllipticE[(c - 
Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(d*Sqrt[(a + b*Sin 
[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/ 
(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c + d*x]])) 
/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3135

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

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

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

rule 3231

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(655\) vs. \(2(156)=312\).

Time = 0.42 (sec) , antiderivative size = 656, normalized size of antiderivative = 3.93

method	result	size
default	\(\frac {2 a^{3} \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right )+\frac {2 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b}{3}-2 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{2} a -\frac {2 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticF}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}}{3}-\frac {8 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}}{3}+\frac {8 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {b \left (\sin \left (d x +c \right )-1\right )}{a +b}}\, \sqrt {-\frac {b \left (1+\sin \left (d x +c \right )\right )}{a -b}}\, \operatorname {EllipticE}\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}}{3}+\frac {2 b^{3} \sin \left (d x +c \right )^{3}}{3}+\frac {2 a \,b^{2} \sin \left (d x +c \right )^{2}}{3}-\frac {2 b^{3} \sin \left (d x +c \right )}{3}-\frac {2 a \,b^{2}}{3}}{\sqrt {a +b \sin \left (d x +c \right )}\, b d \cos \left (d x +c \right )}\)	\(656\)

Input:

int((a+b*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

2/3*(3*a^3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)* 
(-b*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),( 
(a-b)/(a+b))^(1/2))+((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b 
))^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b) 
)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b-3*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(s 
in(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticF(((a+b* 
sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^2*a-((a+b*sin(d*x+c))/(a-b 
))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c))/(a-b))^(1/2)*E 
llipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^3-4*((a+b*s 
in(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b*(1+sin(d*x+c)) 
/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2) 
)*a^3+4*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-b*(sin(d*x+c)-1)/(a+b))^(1/2)*(-b 
*(1+sin(d*x+c))/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a- 
b)/(a+b))^(1/2))*a*b^2+b^3*sin(d*x+c)^3+a*b^2*sin(d*x+c)^2-b^3*sin(d*x+c)- 
a*b^2)/b/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 394, normalized size of antiderivative = 2.36 \[ \int (a+b \sin (c+d x))^{3/2} \, dx=-\frac {2 \, {\left (3 \, \sqrt {b \sin \left (d x + c\right ) + a} b^{2} \cos \left (d x + c\right ) + 12 i \, a \sqrt {\frac {1}{2} i \, b} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right )\right ) - 12 i \, a \sqrt {-\frac {1}{2} i \, b} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right ) - {\left (a^{2} + 3 \, b^{2}\right )} \sqrt {\frac {1}{2} i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right ) - {\left (a^{2} + 3 \, b^{2}\right )} \sqrt {-\frac {1}{2} i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right )}}{9 \, b d} \] Input:

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

Output:

-2/9*(3*sqrt(b*sin(d*x + c) + a)*b^2*cos(d*x + c) + 12*I*a*sqrt(1/2*I*b)*b 
*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3 
, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2 
)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b)) - 12*I*a*sq 
rt(-1/2*I*b)*b*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 
 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I 
*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a) 
/b)) - (a^2 + 3*b^2)*sqrt(1/2*I*b)*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2 
)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin( 
d*x + c) - 2*I*a)/b) - (a^2 + 3*b^2)*sqrt(-1/2*I*b)*weierstrassPInverse(-4 
/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x 
 + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b))/(b*d)
 

Sympy [F]

\[ \int (a+b \sin (c+d x))^{3/2} \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx \] Input:

integrate((a+b*sin(d*x+c))**(3/2),x)
 

Output:

Integral((a + b*sin(c + d*x))**(3/2), x)
 

Maxima [F]

\[ \int (a+b \sin (c+d x))^{3/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \] Input:

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

Output:

integrate((b*sin(d*x + c) + a)^(3/2), x)
 

Giac [F]

\[ \int (a+b \sin (c+d x))^{3/2} \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \] Input:

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

Output:

integrate((b*sin(d*x + c) + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \sin (c+d x))^{3/2} \, dx=\int {\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \] Input:

int((a + b*sin(c + d*x))^(3/2),x)
 

Output:

int((a + b*sin(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int (a+b \sin (c+d x))^{3/2} \, dx=\left (\int \sqrt {\sin \left (d x +c \right ) b +a}d x \right ) a +\left (\int \sqrt {\sin \left (d x +c \right ) b +a}\, \sin \left (d x +c \right )d x \right ) b \] Input:

int((a+b*sin(d*x+c))^(3/2),x)
 

Output:

int(sqrt(sin(c + d*x)*b + a),x)*a + int(sqrt(sin(c + d*x)*b + a)*sin(c + d 
*x),x)*b