3.2.3 \(\int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) [103]

3.2.3.1 Optimal result
3.2.3.2 Mathematica [A] (verified)
3.2.3.3 Rubi [A] (warning: unable to verify)
3.2.3.4 Maple [B] (warning: unable to verify)
3.2.3.5 Fricas [C] (verification not implemented)
3.2.3.6 Sympy [F]
3.2.3.7 Maxima [A] (verification not implemented)
3.2.3.8 Giac [A] (verification not implemented)
3.2.3.9 Mupad [F(-1)]

3.2.3.1 Optimal result

Integrand size = 21, antiderivative size = 257 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b d^{5/2}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b d^{5/2}}-\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b d^{5/2}}+\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b d^{5/2}}+\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{16 b d^3}-\frac {\cos ^4(a+b x) \sqrt {d \tan (a+b x)}}{4 b d^3} \]

output
-3/64*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b/d^(5/2)*2^(1/2)+3/6 
4*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b/d^(5/2)*2^(1/2)-3/128*l 
n(d^(1/2)-2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b/d^(5/2)*2^(1/ 
2)+3/128*ln(d^(1/2)+2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b/d^( 
5/2)*2^(1/2)+1/16*cos(b*x+a)^2*(d*tan(b*x+a))^(1/2)/b/d^3-1/4*cos(b*x+a)^4 
*(d*tan(b*x+a))^(1/2)/b/d^3
 
3.2.3.2 Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.48 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {\csc (a+b x) \left (\sin (a+b x)+3 \arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}-3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}+2 \sin (3 (a+b x))+\sin (5 (a+b x))\right ) \sqrt {d \tan (a+b x)}}{64 b d^3} \]

input
Integrate[Sin[a + b*x]^4/(d*Tan[a + b*x])^(5/2),x]
 
output
-1/64*(Csc[a + b*x]*(Sin[a + b*x] + 3*ArcSin[Cos[a + b*x] - Sin[a + b*x]]* 
Sqrt[Sin[2*(a + b*x)]] - 3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a 
 + b*x)]]]*Sqrt[Sin[2*(a + b*x)]] + 2*Sin[3*(a + b*x)] + Sin[5*(a + b*x)]) 
*Sqrt[d*Tan[a + b*x]])/(b*d^3)
 
3.2.3.3 Rubi [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3071, 252, 253, 266, 755, 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.

\(\displaystyle \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (a+b x)^4}{(d \tan (a+b x))^{5/2}}dx\)

\(\Big \downarrow \) 3071

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

\(\Big \downarrow \) 252

\(\displaystyle \frac {d \left (\frac {1}{8} \int \frac {1}{\sqrt {d \tan (a+b x)} \left (\tan ^2(a+b x) d^2+d^2\right )^2}d(d \tan (a+b x))-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \int \frac {1}{\sqrt {d \tan (a+b x)} \left (\tan ^2(a+b x) d^2+d^2\right )}d(d \tan (a+b x))}{4 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \int \frac {1}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(a+b x)+d}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {d \left (\frac {1}{8} \left (\frac {3 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{2 d^2}+\frac {\sqrt {d \tan (a+b x)}}{2 d^2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {\sqrt {d \tan (a+b x)}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

input
Int[Sin[a + b*x]^4/(d*Tan[a + b*x])^(5/2),x]
 
output
(d*(-1/4*Sqrt[d*Tan[a + b*x]]/(d^2 + d^2*Tan[a + b*x]^2)^2 + ((3*((-(ArcTa 
n[1 - Sqrt[2]*Sqrt[d]*Tan[a + b*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2 
]*Sqrt[d]*Tan[a + b*x]]/(Sqrt[2]*Sqrt[d]))/(2*d) + (-1/2*Log[d - Sqrt[2]*d 
^(3/2)*Tan[a + b*x] + d^2*Tan[a + b*x]^2]/(Sqrt[2]*Sqrt[d]) + Log[d + Sqrt 
[2]*d^(3/2)*Tan[a + b*x] + d^2*Tan[a + b*x]^2]/(2*Sqrt[2]*Sqrt[d]))/(2*d)) 
)/(2*d^2) + Sqrt[d*Tan[a + b*x]]/(2*d^2*(d^2 + d^2*Tan[a + b*x]^2)))/8))/b
 

3.2.3.3.1 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 252
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 253
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   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 3071
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S 
ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f)   Subst[I 
nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], 
x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
 
3.2.3.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(604\) vs. \(2(197)=394\).

Time = 15.09 (sec) , antiderivative size = 605, normalized size of antiderivative = 2.35

method result size
default \(\frac {\csc \left (b x +a \right ) \left (-1+\cos \left (b x +a \right )\right ) \left (16 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \left (\cos ^{4}\left (b x +a \right )\right )+16 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \left (\cos ^{3}\left (b x +a \right )\right )-4 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \left (\cos ^{2}\left (b x +a \right )\right )-4 \cos \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+6 \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\cos \left (b x +a \right )-1}{-1+\cos \left (b x +a \right )}\right )+6 \arctan \left (\frac {\sin \left (b x +a \right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-\cos \left (b x +a \right )+1}{-1+\cos \left (b x +a \right )}\right )-3 \ln \left (\frac {2 \sin \left (b x +a \right ) \sqrt {-\left (\cot ^{3}\left (b x +a \right )\right )+3 \left (\cot ^{2}\left (b x +a \right )\right ) \csc \left (b x +a \right )-3 \cot \left (b x +a \right ) \left (\csc ^{2}\left (b x +a \right )\right )+\csc ^{3}\left (b x +a \right )+\cot \left (b x +a \right )-\csc \left (b x +a \right )}-\cot \left (b x +a \right ) \cos \left (b x +a \right )+2 \cot \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}{-1+\cos \left (b x +a \right )}\right )+3 \ln \left (-\frac {\cot \left (b x +a \right ) \cos \left (b x +a \right )-2 \cot \left (b x +a \right )+2 \sin \left (b x +a \right ) \sqrt {-\left (\cot ^{3}\left (b x +a \right )\right )+3 \left (\cot ^{2}\left (b x +a \right )\right ) \csc \left (b x +a \right )-3 \cot \left (b x +a \right ) \left (\csc ^{2}\left (b x +a \right )\right )+\csc ^{3}\left (b x +a \right )+\cot \left (b x +a \right )-\csc \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}{-1+\cos \left (b x +a \right )}\right )\right ) \sqrt {2}}{128 b \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {d \tan \left (b x +a \right )}\, d^{2}}\) \(605\)

input
int(sin(b*x+a)^4/(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 
output
1/128/b*csc(b*x+a)*(-1+cos(b*x+a))*(16*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a) 
+1)^2)^(1/2)*2^(1/2)*cos(b*x+a)^4+16*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1 
)^2)^(1/2)*2^(1/2)*cos(b*x+a)^3-4*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2 
)^(1/2)*2^(1/2)*cos(b*x+a)^2-4*cos(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/ 
(cos(b*x+a)+1)^2)^(1/2)+6*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+ 
a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))+6*arctan((sin(b* 
x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1) 
/(-1+cos(b*x+a)))-3*ln((2*sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x 
+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-co 
t(b*x+a)*cos(b*x+a)+2*cot(b*x+a)+2*cos(b*x+a)+sin(b*x+a)-csc(b*x+a)-2)/(-1 
+cos(b*x+a)))+3*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-cot 
(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3 
+cot(b*x+a)-csc(b*x+a))^(1/2)-2*cos(b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(-1+co 
s(b*x+a))))/(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)/(d*tan(b*x+a)) 
^(1/2)/d^2*2^(1/2)
 
3.2.3.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 956, normalized size of antiderivative = 3.72 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\text {Too large to display} \]

input
integrate(sin(b*x+a)^4/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")
 
output
1/256*(3*b*d^3*(-1/(b^4*d^10))^(1/4)*log(2*b^2*d^5*sqrt(-1/(b^4*d^10))*cos 
(b*x + a)*sin(b*x + a) - 2*cos(b*x + a)^2 + 2*(b^3*d^7*(-1/(b^4*d^10))^(3/ 
4)*cos(b*x + a)^2 + b*d^2*(-1/(b^4*d^10))^(1/4)*cos(b*x + a)*sin(b*x + a)) 
*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1) - 3*b*d^3*(-1/(b^4*d^10))^(1/4)*lo 
g(2*b^2*d^5*sqrt(-1/(b^4*d^10))*cos(b*x + a)*sin(b*x + a) - 2*cos(b*x + a) 
^2 - 2*(b^3*d^7*(-1/(b^4*d^10))^(3/4)*cos(b*x + a)^2 + b*d^2*(-1/(b^4*d^10 
))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1) 
 + 3*I*b*d^3*(-1/(b^4*d^10))^(1/4)*log(-2*b^2*d^5*sqrt(-1/(b^4*d^10))*cos( 
b*x + a)*sin(b*x + a) - 2*cos(b*x + a)^2 - 2*(I*b^3*d^7*(-1/(b^4*d^10))^(3 
/4)*cos(b*x + a)^2 - I*b*d^2*(-1/(b^4*d^10))^(1/4)*cos(b*x + a)*sin(b*x + 
a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1) - 3*I*b*d^3*(-1/(b^4*d^10))^(1/ 
4)*log(-2*b^2*d^5*sqrt(-1/(b^4*d^10))*cos(b*x + a)*sin(b*x + a) - 2*cos(b* 
x + a)^2 - 2*(-I*b^3*d^7*(-1/(b^4*d^10))^(3/4)*cos(b*x + a)^2 + I*b*d^2*(- 
1/(b^4*d^10))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x 
 + a)) + 1) + 3*b*d^3*(-1/(b^4*d^10))^(1/4)*log(2*(b^3*d^7*(-1/(b^4*d^10)) 
^(3/4)*cos(b*x + a)^2 - b*d^2*(-1/(b^4*d^10))^(1/4)*cos(b*x + a)*sin(b*x + 
 a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 1) - 3*b*d^3*(-1/(b^4*d^10))^(1/4 
)*log(-2*(b^3*d^7*(-1/(b^4*d^10))^(3/4)*cos(b*x + a)^2 - b*d^2*(-1/(b^4*d^ 
10))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 
1) + 3*I*b*d^3*(-1/(b^4*d^10))^(1/4)*log(-2*(I*b^3*d^7*(-1/(b^4*d^10))^...
 
3.2.3.6 Sympy [F]

\[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int \frac {\sin ^{4}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]

input
integrate(sin(b*x+a)**4/(d*tan(b*x+a))**(5/2),x)
 
output
Integral(sin(a + b*x)**4/(d*tan(a + b*x))**(5/2), x)
 
3.2.3.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.85 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {6 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 6 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 3 \, \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 3 \, \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) + \frac {8 \, {\left (\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{4} - 3 \, \sqrt {d \tan \left (b x + a\right )} d^{6}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{128 \, b d^{5}} \]

input
integrate(sin(b*x+a)^4/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")
 
output
1/128*(6*sqrt(2)*d^(5/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*ta 
n(b*x + a)))/sqrt(d)) + 6*sqrt(2)*d^(5/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqr 
t(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 3*sqrt(2)*d^(5/2)*log(d*tan(b*x 
+ a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - 3*sqrt(2)*d^(5/2)*log(d 
*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) + 8*((d*tan(b*x 
+ a))^(5/2)*d^4 - 3*sqrt(d*tan(b*x + a))*d^6)/(d^4*tan(b*x + a)^4 + 2*d^4* 
tan(b*x + a)^2 + d^4))/(b*d^5)
 
3.2.3.8 Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.96 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d^{3}} + \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d^{3}} + \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d^{3}} - \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d^{3}} + \frac {\sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )^{2} - 3 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{16 \, {\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b d} \]

input
integrate(sin(b*x+a)^4/(d*tan(b*x+a))^(5/2),x, algorithm="giac")
 
output
3/64*sqrt(2)*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqr 
t(d*tan(b*x + a)))/sqrt(abs(d)))/(b*d^3) + 3/64*sqrt(2)*sqrt(abs(d))*arcta 
n(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d) 
))/(b*d^3) + 3/128*sqrt(2)*sqrt(abs(d))*log(d*tan(b*x + a) + sqrt(2)*sqrt( 
d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d^3) - 3/128*sqrt(2)*sqrt(abs(d) 
)*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d)) 
/(b*d^3) + 1/16*(sqrt(d*tan(b*x + a))*d^2*tan(b*x + a)^2 - 3*sqrt(d*tan(b* 
x + a))*d^2)/((d^2*tan(b*x + a)^2 + d^2)^2*b*d)
 
3.2.3.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^4}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}} \,d x \]

input
int(sin(a + b*x)^4/(d*tan(a + b*x))^(5/2),x)
 
output
int(sin(a + b*x)^4/(d*tan(a + b*x))^(5/2), x)