\(\int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx\) [267]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\frac {c d \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}{6 b}-\frac {c (d \cos (a+b x))^{5/2} \sqrt {c \sin (a+b x)}}{3 b d}+\frac {c^2 d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{12 b \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \] Output:

1/6*c*d*(d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(1/2)/b-1/3*c*(d*cos(b*x+a))^( 
5/2)*(c*sin(b*x+a))^(1/2)/b/d+1/12*c^2*d^2*InverseJacobiAM(a-1/4*Pi+b*x,2^ 
(1/2))*sin(2*b*x+2*a)^(1/2)/b/(d*cos(b*x+a))^(1/2)/(c*sin(b*x+a))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.54 \[ \int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^{3/2} \, dx=\frac {2 c d \sqrt {d \cos (a+b x)} \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4},\frac {9}{4},\sin ^2(a+b x)\right ) \sqrt {c \sin (a+b x)} \tan ^2(a+b x)}{5 b} \] Input:

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

Output:

(2*c*d*Sqrt[d*Cos[a + b*x]]*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[-1/4, 
 5/4, 9/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a + b*x]]*Tan[a + b*x]^2)/(5*b)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3048, 3042, 3049, 3042, 3053, 3042, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3048

\(\displaystyle \frac {1}{6} c^2 \int \frac {(d \cos (a+b x))^{3/2}}{\sqrt {c \sin (a+b x)}}dx-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} c^2 \int \frac {(d \cos (a+b x))^{3/2}}{\sqrt {c \sin (a+b x)}}dx-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\)

\(\Big \downarrow \) 3049

\(\displaystyle \frac {1}{6} c^2 \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} c^2 \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}dx+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\)

\(\Big \downarrow \) 3053

\(\displaystyle \frac {1}{6} c^2 \left (\frac {d^2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} c^2 \left (\frac {d^2 \sqrt {\sin (2 a+2 b x)} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx}{2 \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{6} c^2 \left (\frac {d^2 \sqrt {\sin (2 a+2 b x)} \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{2 b \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}+\frac {d \sqrt {c \sin (a+b x)} \sqrt {d \cos (a+b x)}}{b c}\right )-\frac {c \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{5/2}}{3 b d}\)

Input:

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

Output:

-1/3*(c*(d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]])/(b*d) + (c^2*((d*Sqrt 
[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])/(b*c) + (d^2*EllipticF[a - Pi/4 + b 
*x, 2]*Sqrt[Sin[2*a + 2*b*x]])/(2*b*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b* 
x]])))/6
 

Defintions of rubi rules used

rule 3042

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

rule 3048

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 
1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Cos[e + f*x])^n 
*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3049

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[a*(b*Sin[e + f*x])^(n + 1)*((a*Cos[e + f*x])^(m - 1)/ 
(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Sin[e + f*x])^n*(a 
*Cos[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && 
 NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3053

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ 
)]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b 
*Cos[e + f*x]])   Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f 
}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 162.14 (sec) , antiderivative size = 993, normalized size of antiderivative = 7.58

method	result	size
default	\(\text {Expression too large to display}\)	\(993\)

Input:

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

Output:

-1/384/b*c*d*(sin(b*x+a)*cos(b*x+a)*(16*cos(b*x+a)^2-8)+(3*cos(b*x+a)+3)*( 
-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-(cos(b*x+a)*cot(b*x+a 
)-2*cot(b*x+a)+2*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sin(b*x 
+a)-2*cos(b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(cos(b*x+a)-1))+(-3*cos(b*x+a)-3 
)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-(cos(b*x+a)*cot(b* 
x+a)-2*cot(b*x+a)-2*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sin( 
b*x+a)-2*cos(b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(cos(b*x+a)-1))+(6*cos(b*x+a) 
+6)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan(((-2*sin(b*x+ 
a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sin(b*x+a)+cos(b*x+a)-1)/(cos(b*x+a) 
-1))+(-6*cos(b*x+a)-6)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*a 
rctan((-(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sin(b*x+a)+cos(b 
*x+a)-1)/(cos(b*x+a)-1))+(8*cos(b*x+a)+8)*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2) 
*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*Ellipti 
cF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))+(-6*cos(b*x+a)-6)*(-cot(b 
*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)- 
csc(b*x+a))^(1/2)*EllipticPi((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2-1/2*I,1/ 
2*2^(1/2))+(-6*cos(b*x+a)-6)*(-cot(b*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a 
)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((-cot(b*x 
+a)+csc(b*x+a)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))+I*(-6*cos(b*x+a)-6)*(-cot(b 
*x+a)+csc(b*x+a)+1)^(1/2)*(2*cot(b*x+a)-2*csc(b*x+a)+2)^(1/2)*(cot(b*x+...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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