\(\int \frac {\sin ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\) [84]

Optimal result
Mathematica [A] (verified)
Rubi [A] (warning: unable to verify)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

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

-5/64*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b/d^(1/2)+5/6 
4*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b/d^(1/2)+5/64*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^(1/2)-5/16*cos(b*x+a)^2*(d*tan(b*x+a))^(1/2)/b/d-1/4*cos(b*x+a)^4*(d*ta 
n(b*x+a))^(5/2)/b/d^3
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.60 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\frac {\sec (a+b x) \left (-7 \sin (a+b x)-5 \arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}+5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}-6 \sin (3 (a+b x))+\sin (5 (a+b x))\right )}{64 b \sqrt {d \tan (a+b x)}} \] Input:

Integrate[Sin[a + b*x]^4/Sqrt[d*Tan[a + b*x]],x]
 

Output:

(Sec[a + b*x]*(-7*Sin[a + b*x] - 5*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Sqr 
t[Sin[2*(a + b*x)]] + 5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + 
b*x)]]]*Sqrt[Sin[2*(a + b*x)]] - 6*Sin[3*(a + b*x)] + Sin[5*(a + b*x)]))/( 
64*b*Sqrt[d*Tan[a + b*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.26, 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, 252, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3071

\(\displaystyle \frac {d \int \frac {(d \tan (a+b x))^{7/2}}{\left (\tan ^2(a+b x) d^2+d^2\right )^3}d(d \tan (a+b x))}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {d \left (\frac {5}{8} \int \frac {(d \tan (a+b x))^{3/2}}{\left (\tan ^2(a+b x) d^2+d^2\right )^2}d(d \tan (a+b x))-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{4} \int \frac {1}{\sqrt {d \tan (a+b x)} \left (\tan ^2(a+b x) d^2+d^2\right )}d(d \tan (a+b x))-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \int \frac {1}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(a+b x)+d}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {d \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )-\frac {\sqrt {d \tan (a+b x)}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{5/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

Input:

Int[Sin[a + b*x]^4/Sqrt[d*Tan[a + b*x]],x]
 

Output:

(d*(-1/4*(d*Tan[a + b*x])^(5/2)/(d^2 + d^2*Tan[a + b*x]^2)^2 + (5*(((-(Arc 
Tan[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 + Sq 
rt[2]*d^(3/2)*Tan[a + b*x] + d^2*Tan[a + b*x]^2]/(2*Sqrt[2]*Sqrt[d]))/(2*d 
))/2 - Sqrt[d*Tan[a + b*x]]/(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 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. \(547\) vs. \(2(158)=316\).

Time = 7.86 (sec) , antiderivative size = 548, normalized size of antiderivative = 2.69

method result size
default \(\frac {\left (-5 \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 )+5 \ln \left (\frac {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}}}-\cot \left (b x +a \right ) \cos \left (b x +a \right )+\sin \left (b x +a \right )+2 \cos \left (b x +a \right )-\csc \left (b x +a \right )+2 \cot \left (b x +a \right )-2}{-1+\cos \left (b x +a \right )}\right ) \sin \left (b x +a \right )-10 \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 )-10 \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 (-80 \cos \left (b x +a \right )^{2}+\left (80 \sin \left (b x +a \right )-80\right ) \cos \left (b x +a \right )+80 \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 (24 \cos \left (b x +a \right )^{2}-29\right ) \left (\cos \left (b x +a \right )+1\right ) \sin \left (b x +a \right )-80 \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 ) \sqrt {d \tan \left (b x +a \right )}}\) \(548\)

Input:

int(sin(b*x+a)^4/(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/128/b*(-5*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)+5*ln((2*sin(b*x+a)*(-2*sin(b*x+a)*cos(b*x 
+a)/(cos(b*x+a)+1)^2)^(1/2)-cot(b*x+a)*cos(b*x+a)+sin(b*x+a)+2*cos(b*x+a)- 
csc(b*x+a)+2*cot(b*x+a)-2)/(-1+cos(b*x+a)))*sin(b*x+a)-10*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)-10*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)*(- 
80*cos(b*x+a)^2+(80*sin(b*x+a)-80)*cos(b*x+a)+80*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*(24*cos(b* 
x+a)^2-29)*(cos(b*x+a)+1)*sin(b*x+a)-80*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*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 = 0.14 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.10 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\frac {10 \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )}}{2 \, \sqrt {d} \sin \left (b x + a\right )}\right ) - 5 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} - 2}{2 \, {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1\right )}}\right ) - 5 \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {2 \, \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} - 2}{2 \, {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1\right )}}\right ) + 5 \, \sqrt {2} \sqrt {d} \log \left (4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac {2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} + 1\right ) - 5 \, \sqrt {2} \sqrt {d} \log \left (4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \frac {2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{\sqrt {d}} + 1\right ) + 16 \, {\left (4 \, \cos \left (b x + a\right )^{4} - 9 \, \cos \left (b x + a\right )^{2}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{256 \, b d} \] Input:

integrate(sin(b*x+a)^4/(d*tan(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

1/256*(10*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))) - 5*sqrt(2)*sq 
rt(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)) - 5*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))/ 
sqrt(d) - 2)/(cos(b*x + a)^2 + cos(b*x + a)*sin(b*x + a) - 1)) + 5*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) - 5 
*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 - 9*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b* 
x + a)))/(b*d)
 

Sympy [F]

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

integrate(sin(b*x+a)**4/(d*tan(b*x+a))**(1/2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\frac {10 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 10 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 5 \, \sqrt {2} d^{\frac {9}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 5 \, \sqrt {2} d^{\frac {9}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \frac {8 \, {\left (9 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{6} + 5 \, \sqrt {d \tan \left (b x + a\right )} d^{8}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{128 \, b d^{5}} \] Input:

integrate(sin(b*x+a)^4/(d*tan(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

1/128*(10*sqrt(2)*d^(9/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*t 
an(b*x + a)))/sqrt(d)) + 10*sqrt(2)*d^(9/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*s 
qrt(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 5*sqrt(2)*d^(9/2)*log(d*tan(b* 
x + a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - 5*sqrt(2)*d^(9/2)*log 
(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - 8*(9*(d*tan( 
b*x + a))^(5/2)*d^6 + 5*sqrt(d*tan(b*x + a))*d^8)/(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 = 0.14 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.21 \[ \int \frac {\sin ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx=\frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d} + \frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d} + \frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d} - \frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d} - \frac {9 \, \sqrt {d \tan \left (b x + a\right )} d^{3} \tan \left (b x + a\right )^{2} + 5 \, \sqrt {d \tan \left (b x + a\right )} d^{3}}{16 \, {\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b} \] Input:

integrate(sin(b*x+a)^4/(d*tan(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

5/64*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) + 5/64*sqrt(2)*sqrt(abs(d))*arctan( 
-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d))) 
/(b*d) + 5/128*sqrt(2)*sqrt(abs(d))*log(d*tan(b*x + a) + sqrt(2)*sqrt(d*ta 
n(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d) - 5/128*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) 
 - 1/16*(9*sqrt(d*tan(b*x + a))*d^3*tan(b*x + a)^2 + 5*sqrt(d*tan(b*x + a) 
)*d^3)/((d^2*tan(b*x + a)^2 + d^2)^2*b)
 

Mupad [F(-1)]

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

int(sin(a + b*x)^4/(d*tan(a + b*x))^(1/2),x)
 

Output:

int(sin(a + b*x)^4/(d*tan(a + b*x))^(1/2), x)
 

Reduce [F]

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

int(sin(b*x+a)^4/(d*tan(b*x+a))^(1/2),x)
 

Output:

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