\(\int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx\) [282]

Optimal result

Integrand size = 25, antiderivative size = 245 \[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {3 c^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {3 c^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{4 \sqrt {2} b \sqrt {d}}-\frac {3 c^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)} \left (\sqrt {c}+\sqrt {c} \tan (a+b x)\right )}\right )}{4 \sqrt {2} b \sqrt {d}}-\frac {c \sqrt {d \cos (a+b x)} (c \sin (a+b x))^{3/2}}{2 b d} \] Output:

-3/8*c^(5/2)*arctan(1-2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/c^(1/2)/(d*cos( 
b*x+a))^(1/2))*2^(1/2)/b/d^(1/2)+3/8*c^(5/2)*arctan(1+2^(1/2)*d^(1/2)*(c*s 
in(b*x+a))^(1/2)/c^(1/2)/(d*cos(b*x+a))^(1/2))*2^(1/2)/b/d^(1/2)-3/8*c^(5/ 
2)*arctanh(2^(1/2)*d^(1/2)*(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(1/2)/(c^(1 
/2)+c^(1/2)*tan(b*x+a)))*2^(1/2)/b/d^(1/2)-1/2*c*(d*cos(b*x+a))^(1/2)*(c*s 
in(b*x+a))^(3/2)/b/d
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.13 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.27 \[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=\frac {2 \cos ^2(a+b x)^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{4},\frac {11}{4},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{5/2} \tan (a+b x)}{7 b \sqrt {d \cos (a+b x)}} \] Input:

Integrate[(c*Sin[a + b*x])^(5/2)/Sqrt[d*Cos[a + b*x]],x]
 

Output:

(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[3/4, 7/4, 11/4, Sin[a + b*x]^2 
]*(c*Sin[a + b*x])^(5/2)*Tan[a + b*x])/(7*b*Sqrt[d*Cos[a + b*x]])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.33, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3048, 3042, 3054, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3048

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3054

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 c^3 d \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}+1\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {c} \sqrt {d \cos (a+b x)}}\right )}{\sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}-\frac {\frac {\log \left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+c \tan (a+b x)+c\right )}{2 \sqrt {2} \sqrt {c} \sqrt {d}}-\frac {\log \left (-\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {c \sin (a+b x)}}{\sqrt {d \cos (a+b x)}}+c \tan (a+b x)+c\right )}{2 \sqrt {2} \sqrt {c} \sqrt {d}}}{2 d}\right )}{2 b}-\frac {c (c \sin (a+b x))^{3/2} \sqrt {d \cos (a+b x)}}{2 b d}\)

Input:

Int[(c*Sin[a + b*x])^(5/2)/Sqrt[d*Cos[a + b*x]],x]
 

Output:

(3*c^3*d*((-(ArcTan[1 - (Sqrt[2]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/(Sqrt[c]*Sq 
rt[d*Cos[a + b*x]])]/(Sqrt[2]*Sqrt[c]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt 
[d]*Sqrt[c*Sin[a + b*x]])/(Sqrt[c]*Sqrt[d*Cos[a + b*x]])]/(Sqrt[2]*Sqrt[c] 
*Sqrt[d]))/(2*d) - (-1/2*Log[c - (Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[c*Sin[a + b 
*x]])/Sqrt[d*Cos[a + b*x]] + c*Tan[a + b*x]]/(Sqrt[2]*Sqrt[c]*Sqrt[d]) + L 
og[c + (Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[c*Sin[a + b*x]])/Sqrt[d*Cos[a + b*x]] 
 + c*Tan[a + b*x]]/(2*Sqrt[2]*Sqrt[c]*Sqrt[d]))/(2*d)))/(2*b) - (c*Sqrt[d* 
Cos[a + b*x]]*(c*Sin[a + b*x])^(3/2))/(2*b*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

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

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

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 3054

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(436\) vs. \(2(183)=366\).

Time = 6.98 (sec) , antiderivative size = 437, normalized size of antiderivative = 1.78

method	result	size
default	\(-\frac {\sqrt {2}\, c^{2} \left (-3 \ln \left (-\frac {\cos \left (b x +a \right ) \cot \left (b x +a \right )-2 \cot \left (b x +a \right )-2 \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )-2 \cos \left (b x +a \right )-\sin \left (b x +a \right )+\csc \left (b x +a \right )+2}{\cos \left (b x +a \right )-1}\right )+3 \ln \left (-\frac {\cos \left (b x +a \right ) \cot \left (b x +a \right )-2 \cot \left (b x +a \right )+2 \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )-2 \cos \left (b x +a \right )-\sin \left (b x +a \right )+\csc \left (b x +a \right )+2}{\cos \left (b x +a \right )-1}\right )+6 \arctan \left (\frac {-\sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )+\cos \left (b x +a \right )-1}{\cos \left (b x +a \right )-1}\right )-6 \arctan \left (\frac {\sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sin \left (b x +a \right )+\cos \left (b x +a \right )-1}{\cos \left (b x +a \right )-1}\right )+\left (4 \cos \left (b x +a \right )+4\right ) \sin \left (b x +a \right ) \sqrt {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\right ) \sqrt {c \sin \left (b x +a \right )}\, \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3} \sec \left (\frac {b x}{2}+\frac {a}{2}\right )^{5} \csc \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}{256 b \sqrt {d \cos \left (b x +a \right )}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(437\)

Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/256/b*2^(1/2)*c^2*(-3*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*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*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 
*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))+(4*cos(b*x+a)+4)*sin(b*x+a)*(-2*sin(b*x+a)*cos(b 
*x+a)/(cos(b*x+a)+1)^2)^(1/2))*(c*sin(b*x+a))^(1/2)*cos(b*x+a)*sin(b*x+a)^ 
3/(d*cos(b*x+a))^(1/2)/(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*sec 
(1/2*b*x+1/2*a)^5*csc(1/2*b*x+1/2*a)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (183) = 366\).

Time = 0.15 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.18 \[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {3 \, \sqrt {2} c^{2} d \sqrt {\frac {c}{d}} \arctan \left (\frac {2 \, c \cos \left (b x + a\right )^{3} - 2 \, c \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + \sqrt {2} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sqrt {\frac {c}{d}} - 2 \, c \cos \left (b x + a\right )}{2 \, {\left (c \cos \left (b x + a\right )^{3} + c \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - c \cos \left (b x + a\right )\right )}}\right ) + 3 \, \sqrt {2} c^{2} d \sqrt {\frac {c}{d}} \arctan \left (-\frac {2 \, c \cos \left (b x + a\right )^{3} - 2 \, c \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - \sqrt {2} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sqrt {\frac {c}{d}} - 2 \, c \cos \left (b x + a\right )}{2 \, {\left (c \cos \left (b x + a\right )^{3} + c \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) - c \cos \left (b x + a\right )\right )}}\right ) - 6 \, \sqrt {2} c^{2} d \sqrt {\frac {c}{d}} \arctan \left (-\frac {\sqrt {2} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sqrt {\frac {c}{d}} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )}}{2 \, c \cos \left (b x + a\right ) \sin \left (b x + a\right )}\right ) + 3 \, \sqrt {2} c^{2} d \sqrt {\frac {c}{d}} \log \left (2 \, \sqrt {2} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sqrt {\frac {c}{d}} {\left (\cos \left (b x + a\right ) + \sin \left (b x + a\right )\right )} + 4 \, c \cos \left (b x + a\right ) \sin \left (b x + a\right ) + c\right ) - 3 \, \sqrt {2} c^{2} d \sqrt {\frac {c}{d}} \log \left (-2 \, \sqrt {2} \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sqrt {\frac {c}{d}} {\left (\cos \left (b x + a\right ) + \sin \left (b x + a\right )\right )} + 4 \, c \cos \left (b x + a\right ) \sin \left (b x + a\right ) + c\right ) + 16 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} c^{2} \sin \left (b x + a\right )}{32 \, b d} \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

-1/32*(3*sqrt(2)*c^2*d*sqrt(c/d)*arctan(1/2*(2*c*cos(b*x + a)^3 - 2*c*cos( 
b*x + a)^2*sin(b*x + a) + sqrt(2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a) 
)*sqrt(c/d) - 2*c*cos(b*x + a))/(c*cos(b*x + a)^3 + c*cos(b*x + a)^2*sin(b 
*x + a) - c*cos(b*x + a))) + 3*sqrt(2)*c^2*d*sqrt(c/d)*arctan(-1/2*(2*c*co 
s(b*x + a)^3 - 2*c*cos(b*x + a)^2*sin(b*x + a) - sqrt(2)*sqrt(d*cos(b*x + 
a))*sqrt(c*sin(b*x + a))*sqrt(c/d) - 2*c*cos(b*x + a))/(c*cos(b*x + a)^3 + 
 c*cos(b*x + a)^2*sin(b*x + a) - c*cos(b*x + a))) - 6*sqrt(2)*c^2*d*sqrt(c 
/d)*arctan(-1/2*sqrt(2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*sqrt(c/d 
)*(cos(b*x + a) - sin(b*x + a))/(c*cos(b*x + a)*sin(b*x + a))) + 3*sqrt(2) 
*c^2*d*sqrt(c/d)*log(2*sqrt(2)*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*s 
qrt(c/d)*(cos(b*x + a) + sin(b*x + a)) + 4*c*cos(b*x + a)*sin(b*x + a) + c 
) - 3*sqrt(2)*c^2*d*sqrt(c/d)*log(-2*sqrt(2)*sqrt(d*cos(b*x + a))*sqrt(c*s 
in(b*x + a))*sqrt(c/d)*(cos(b*x + a) + sin(b*x + a)) + 4*c*cos(b*x + a)*si 
n(b*x + a) + c) + 16*sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*c^2*sin(b*x 
 + a))/(b*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=\text {Timed out} \] Input:

integrate((c*sin(b*x+a))**(5/2)/(d*cos(b*x+a))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\sqrt {d \cos \left (b x + a\right )}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

integrate((c*sin(b*x + a))^(5/2)/sqrt(d*cos(b*x + a)), x)
 

Giac [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=\int { \frac {\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}{\sqrt {d \cos \left (b x + a\right )}} \,d x } \] Input:

integrate((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

integrate((c*sin(b*x + a))^(5/2)/sqrt(d*cos(b*x + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=\int \frac {{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}}{\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \] Input:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(1/2),x)
 

Output:

int((c*sin(a + b*x))^(5/2)/(d*cos(a + b*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {(c \sin (a+b x))^{5/2}}{\sqrt {d \cos (a+b x)}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \left (\int \frac {\sqrt {\sin \left (b x +a \right )}\, \sqrt {\cos \left (b x +a \right )}\, \sin \left (b x +a \right )^{2}}{\cos \left (b x +a \right )}d x \right ) c^{2}}{d} \] Input:

int((c*sin(b*x+a))^(5/2)/(d*cos(b*x+a))^(1/2),x)
 

Output:

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