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

Optimal result

Integrand size = 21, antiderivative size = 204 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b d^{5/2}}+\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b d^{5/2}}+\frac {3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}+\sqrt {d} \tan (a+b x)}\right )}{32 \sqrt {2} b d^{5/2}}+\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{16 b d^3}-\frac {\cos ^4(a+b x) \sqrt {d \tan (a+b x)}}{4 b d^3} \] Output:

-3/64*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b/d^(5/2)+3/6 
4*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b/d^(5/2)+3/64*ar 
ctanh(2^(1/2)*(d*tan(b*x+a))^(1/2)/(d^(1/2)+d^(1/2)*tan(b*x+a)))*2^(1/2)/b 
/d^(5/2)+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
 

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=-\frac {\csc (a+b x) \left (\sin (a+b x)+3 \arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}-3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}+2 \sin (3 (a+b x))+\sin (5 (a+b x))\right ) \sqrt {d \tan (a+b x)}}{64 b d^3} \] 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)
 

Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.29, 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
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(549\) vs. \(2(158)=316\).

Time = 8.61 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.70

method	result	size
default	\(-\frac {\left (-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 {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-2 \cos \left (b x +a \right )+\csc \left (b x +a \right )-\sin \left (b x +a \right )+2}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \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 {-\frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-2 \cos \left (b x +a \right )+\csc \left (b x +a \right )-\sin \left (b x +a \right )+2}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )-6 \arctan \left (\frac {-\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}}}+\cos \left (b x +a \right )-1}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )+6 \arctan \left (\frac {\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}}}+\cos \left (b x +a \right )-1}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )+\cos \left (b x +a \right ) \left (48 \cos \left (b x +a \right )^{2}+\left (-48 \sin \left (b x +a \right )+48\right ) \cos \left (b x +a \right )-48 \sin \left (b x +a \right )\right ) \sin \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}}}+\cos \left (b x +a \right ) \left (-4 \left (8 \cos \left (b x +a \right )^{2}-11\right ) \left (\cos \left (b x +a \right )+1\right ) \sin \left (b x +a \right )+48 \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )^{2}-1\right ) \left (\cos \left (b x +a \right )+1\right )\right ) \sqrt {2}\, \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\right ) \sqrt {2}}{128 b \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \left (\cos \left (b x +a \right )+1\right ) d^{2} \sqrt {d \tan \left (b x +a \right )}}\)	\(550\)

Input:

int(sin(b*x+a)^4/(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/128/b*(-3*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)-2*sin(b*x+a)*(-2*sin( 
b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)+csc(b*x+a)-sin(b*x+ 
a)+2)/(-1+cos(b*x+a)))*sin(b*x+a)+3*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a 
)+2*sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x 
+a)+csc(b*x+a)-sin(b*x+a)+2)/(-1+cos(b*x+a)))*sin(b*x+a)-6*arctan((-sin(b* 
x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+c 
os(b*x+a)))*sin(b*x+a)+6*arctan((sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos 
(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*sin(b*x+a)+cos(b*x+a)*( 
48*cos(b*x+a)^2+(-48*sin(b*x+a)+48)*cos(b*x+a)-48*sin(b*x+a))*sin(b*x+a)^2 
*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)*(-4*(8*cos(b 
*x+a)^2-11)*(cos(b*x+a)+1)*sin(b*x+a)+48*cos(b*x+a)*(cos(b*x+a)^2-1)*(cos( 
b*x+a)+1))*2^(1/2)*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2))/(-sin( 
b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)/(cos(b*x+a)+1)/d^2/(d*tan(b*x+a) 
)^(1/2)*2^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (158) = 316\).

Time = 0.13 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.10 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {6 \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )}}{2 \, \sqrt {d} \sin \left (b x + a\right )}\right ) - 3 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} - 2}{2 \, {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1\right )}}\right ) - 3 \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} - 2}{2 \, {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1\right )}}\right ) + 3 \, \sqrt {2} \sqrt {d} \log \left (4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac {2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} + 1\right ) - 3 \, \sqrt {2} \sqrt {d} \log \left (4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \frac {2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} + 1\right ) - 16 \, {\left (4 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{256 \, b d^{3}} \] Input:

integrate(sin(b*x+a)^4/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")
 

Output:

1/256*(6*sqrt(2)*sqrt(d)*arctan(-1/2*sqrt(2)*sqrt(d*sin(b*x + a)/cos(b*x + 
 a))*(cos(b*x + a) - sin(b*x + a))/(sqrt(d)*sin(b*x + a))) - 3*sqrt(2)*sqr 
t(d)*arctan(1/2*(2*cos(b*x + a)^2 - 2*cos(b*x + a)*sin(b*x + a) + sqrt(2)* 
sqrt(d*sin(b*x + a)/cos(b*x + a))/sqrt(d) - 2)/(cos(b*x + a)^2 + cos(b*x + 
 a)*sin(b*x + a) - 1)) - 3*sqrt(2)*sqrt(d)*arctan(-1/2*(2*cos(b*x + a)^2 - 
 2*cos(b*x + a)*sin(b*x + a) - sqrt(2)*sqrt(d*sin(b*x + a)/cos(b*x + a))/s 
qrt(d) - 2)/(cos(b*x + a)^2 + cos(b*x + a)*sin(b*x + a) - 1)) + 3*sqrt(2)* 
sqrt(d)*log(4*cos(b*x + a)*sin(b*x + a) + 2*sqrt(2)*(cos(b*x + a)^2 + cos( 
b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))/sqrt(d) + 1) - 3* 
sqrt(2)*sqrt(d)*log(4*cos(b*x + a)*sin(b*x + a) - 2*sqrt(2)*(cos(b*x + a)^ 
2 + cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))/sqrt(d) + 
 1) - 16*(4*cos(b*x + a)^4 - cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + 
 a)))/(b*d^3)
                                                                                    
                                                                                    
 

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)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.07 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {6 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 6 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 3 \, \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 3 \, \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) + \frac {8 \, {\left (\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{4} - 3 \, \sqrt {d \tan \left (b x + a\right )} d^{6}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{128 \, b d^{5}} \] 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)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.22 \[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d^{3}} + \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d^{3}} + \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d^{3}} - \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d^{3}} + \frac {\sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )^{2} - 3 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{16 \, {\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b d} \] 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)
 

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)
 

Reduce [F]

\[ \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (b x +a \right )}\, \sin \left (b x +a \right )^{4}}{\tan \left (b x +a \right )^{3}}d x \right )}{d^{3}} \] Input:

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

Output:

(sqrt(d)*int((sqrt(tan(a + b*x))*sin(a + b*x)**4)/tan(a + b*x)**3,x))/d**3