\(\int \frac {(d \cos (a+b x))^{3/2}}{\sqrt {c \sin (a+b x)}} \, dx\) [292]

Optimal result

Integrand size = 25, antiderivative size = 92 \[ \int \frac {(d \cos (a+b x))^{3/2}}{\sqrt {c \sin (a+b x)}} \, dx=\frac {d \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}}{b c}+\frac {d^2 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)}}{2 b \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \] Output:

d*(d*cos(b*x+a))^(1/2)*(c*sin(b*x+a))^(1/2)/b/c+1/2*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 = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.74 \[ \int \frac {(d \cos (a+b x))^{3/2}}{\sqrt {c \sin (a+b x)}} \, dx=\frac {2 d^2 \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {5}{4},\sin ^2(a+b x)\right ) \tan (a+b x)}{b \sqrt {d \cos (a+b x)} \sqrt {c \sin (a+b x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3049

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3053

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(d*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])/(b*c) + (d^2*EllipticF[a - P 
i/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]])
 

Defintions of rubi rules used

rule 3042

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

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 [A] (verified)

Time = 3.86 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {\sqrt {d \cos \left (b x +a \right )}\, d \left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}\, \sqrt {2 \cot \left (b x +a \right )-2 \csc \left (b x +a \right )+2}\, \sqrt {\cot \left (b x +a \right )-\csc \left (b x +a \right )}\, \operatorname {EllipticF}\left (\sqrt {-\cot \left (b x +a \right )+\csc \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (1+\sec \left (b x +a \right )\right )+2 \sin \left (b x +a \right )\right )}{2 b \sqrt {c \sin \left (b x +a \right )}}\)	\(124\)

Input:

int((d*cos(b*x+a))^(3/2)/(c*sin(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2/b*(d*cos(b*x+a))^(1/2)*d/(c*sin(b*x+a))^(1/2)*((-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 
)*EllipticF((-cot(b*x+a)+csc(b*x+a)+1)^(1/2),1/2*2^(1/2))*(1+sec(b*x+a))+2 
*sin(b*x+a))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((d*cos(b*x+a))**(3/2)/(c*sin(b*x+a))**(1/2),x)
 

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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