\(\int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx\) [74]

Optimal result

Integrand size = 21, antiderivative size = 194 \[ \int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {7 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}-\frac {7 d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}+\frac {7 d^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}+\sqrt {d} \tan (a+b x)}\right )}{4 \sqrt {2} b}+\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d} \] Output:

7/8*d^(5/2)*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b-7/8*d 
^(5/2)*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b+7/8*d^(5/2 
)*arctanh(2^(1/2)*(d*tan(b*x+a))^(1/2)/(d^(1/2)+d^(1/2)*tan(b*x+a)))*2^(1/ 
2)/b+7/6*d*(d*tan(b*x+a))^(3/2)/b-1/2*cos(b*x+a)^2*(d*tan(b*x+a))^(7/2)/b/ 
d
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.65 \[ \int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {d \left (16+12 \cos ^2(a+b x)+21 \arcsin (\cos (a+b x)-\sin (a+b x)) \cot (a+b x) \csc (a+b x) \sqrt {\sin (2 (a+b x))}+21 \cot (a+b x) \csc (a+b x) \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}\right ) (d \tan (a+b x))^{3/2}}{24 b} \] Input:

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

Output:

(d*(16 + 12*Cos[a + b*x]^2 + 21*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Cot[a 
+ b*x]*Csc[a + b*x]*Sqrt[Sin[2*(a + b*x)]] + 21*Cot[a + b*x]*Csc[a + b*x]* 
Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + 
b*x)]])*(d*Tan[a + b*x])^(3/2))/(24*b)
 

Rubi [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.21, 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, 262, 266, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3071

\(\displaystyle \frac {d \int \frac {(d \tan (a+b x))^{9/2}}{\left (\tan ^2(a+b x) d^2+d^2\right )^2}d(d \tan (a+b x))}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {d \left (\frac {7}{4} \int \frac {(d \tan (a+b x))^{5/2}}{\tan ^2(a+b x) d^2+d^2}d(d \tan (a+b x))-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-d^2 \int \frac {\sqrt {d \tan (a+b x)}}{\tan ^2(a+b x) d^2+d^2}d(d \tan (a+b x))\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \int \frac {d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \int \frac {d^2 \tan ^2(a+b x)+d}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}-\frac {1}{2} \int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}\right )\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {d \left (\frac {7}{4} \left (\frac {2}{3} (d \tan (a+b x))^{3/2}-2 d^2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )\right )-\frac {(d \tan (a+b x))^{7/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )}{b}\)

Input:

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

Output:

(d*(-1/2*(d*Tan[a + b*x])^(7/2)/(d^2 + d^2*Tan[a + b*x]^2) + (7*(-2*d^2*(( 
-(ArcTan[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 + (Log[d - Sqrt[2]*d^( 
3/2)*Tan[a + b*x] + d^2*Tan[a + b*x]^2]/(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) + ( 
2*(d*Tan[a + b*x])^(3/2))/3))/4))/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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 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 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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(487\) vs. \(2(148)=296\).

Time = 5.28 (sec) , antiderivative size = 488, normalized size of antiderivative = 2.52

method	result	size
default	\(-\frac {\sqrt {d \tan \left (b x +a \right )}\, \left (21 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \ln \left (2 \cot \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 \csc \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 \cot \left (b x +a \right )+2\right ) \left (\cot \left (b x +a \right )+\csc \left (b x +a \right )\right )+21 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \ln \left (-2 \cot \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 \csc \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 \cot \left (b x +a \right )+2\right ) \left (-\cot \left (b x +a \right )-\csc \left (b x +a \right )\right )+42 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \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 ) \left (-\cot \left (b x +a \right )-\csc \left (b x +a \right )\right )+42 \sqrt {-\frac {\sin \left (b x +a \right ) \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \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 ) \left (-\cot \left (b x +a \right )-\csc \left (b x +a \right )\right )+4 \tan \left (b x +a \right ) \left (-4-3 \cos \left (b x +a \right )^{2}\right ) \sqrt {2}\right ) d^{2} \sqrt {2}}{48 b}\)	\(488\)

Input:

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

Output:

-1/48/b*(d*tan(b*x+a))^(1/2)*(21*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2) 
^(1/2)*ln(2*cot(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+2 
*csc(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cot(b*x+a) 
+2)*(cot(b*x+a)+csc(b*x+a))+21*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^( 
1/2)*ln(-2*cot(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2* 
csc(b*x+a)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-2*cot(b*x+a)+ 
2)*(-cot(b*x+a)-csc(b*x+a))+42*(-sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^( 
1/2)*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)))*(-cot(b*x+a)-csc(b*x+a))+42*(-sin(b*x+a)*co 
s(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*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)))*(-cot(b*x+a)-cs 
c(b*x+a))+4*tan(b*x+a)*(-4-3*cos(b*x+a)^2)*2^(1/2))*d^2*2^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (148) = 296\).

Time = 0.20 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.53 \[ \int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {42 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} \sqrt {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )}{d \cos \left (b x + a\right ) - d \sin \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + 21 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {2 \, d \cos \left (b x + a\right )^{2} - 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \sqrt {2} \sqrt {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} - 2 \, d}{2 \, {\left (d \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - d\right )}}\right ) \cos \left (b x + a\right ) + 21 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {2 \, d \cos \left (b x + a\right )^{2} - 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \sqrt {2} \sqrt {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} - 2 \, d}{2 \, {\left (d \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - d\right )}}\right ) \cos \left (b x + a\right ) + 21 \, \sqrt {2} d^{\frac {5}{2}} \cos \left (b x + a\right ) \log \left (4 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 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 {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} + d\right ) - 21 \, \sqrt {2} d^{\frac {5}{2}} \cos \left (b x + a\right ) \log \left (4 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 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 {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} + d\right ) + 16 \, {\left (3 \, d^{2} \cos \left (b x + a\right )^{2} + 4 \, d^{2}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \sin \left (b x + a\right )}{96 \, b \cos \left (b x + a\right )} \] Input:

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

Output:

1/96*(42*sqrt(2)*d^(5/2)*arctan(-sqrt(2)*sqrt(d)*sqrt(d*sin(b*x + a)/cos(b 
*x + a))*cos(b*x + a)/(d*cos(b*x + a) - d*sin(b*x + a)))*cos(b*x + a) + 21 
*sqrt(2)*d^(5/2)*arctan(1/2*(2*d*cos(b*x + a)^2 - 2*d*cos(b*x + a)*sin(b*x 
 + a) + sqrt(2)*sqrt(d)*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 2*d)/(d*cos(b* 
x + a)^2 + d*cos(b*x + a)*sin(b*x + a) - d))*cos(b*x + a) + 21*sqrt(2)*d^( 
5/2)*arctan(-1/2*(2*d*cos(b*x + a)^2 - 2*d*cos(b*x + a)*sin(b*x + a) - sqr 
t(2)*sqrt(d)*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 2*d)/(d*cos(b*x + a)^2 + 
d*cos(b*x + a)*sin(b*x + a) - d))*cos(b*x + a) + 21*sqrt(2)*d^(5/2)*cos(b* 
x + a)*log(4*d*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)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + d) - 2 
1*sqrt(2)*d^(5/2)*cos(b*x + a)*log(4*d*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)*sqrt(d*sin(b*x + a 
)/cos(b*x + a)) + d) + 16*(3*d^2*cos(b*x + a)^2 + 4*d^2)*sqrt(d*sin(b*x + 
a)/cos(b*x + a))*sin(b*x + a))/(b*cos(b*x + a))
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

integrate(sin(b*x+a)**2*(d*tan(b*x+a))**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.08 \[ \int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {21 \, d^{6} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - \frac {24 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{6}}{d^{2} \tan \left (b x + a\right )^{2} + d^{2}} - 32 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{4}}{48 \, b d^{3}} \] Input:

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

Output:

-1/48*(21*d^6*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*ta 
n(b*x + a)))/sqrt(d))/sqrt(d) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqr 
t(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) - sqrt(2)*log(d*tan(b*x + 
a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d)/sqrt(d) + sqrt(2)*log(d*tan 
(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d)/sqrt(d)) - 24*(d*tan 
(b*x + a))^(3/2)*d^6/(d^2*tan(b*x + a)^2 + d^2) - 32*(d*tan(b*x + a))^(3/2 
)*d^4)/(b*d^3)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.30 \[ \int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {1}{48} \, {\left (\frac {24 \, \sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b} - \frac {42 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d} - \frac {42 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d} + \frac {21 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d} - \frac {21 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d} + \frac {32 \, \sqrt {d \tan \left (b x + a\right )} \tan \left (b x + a\right )}{b}\right )} d^{2} \] Input:

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

Output:

1/48*(24*sqrt(d*tan(b*x + a))*d^2*tan(b*x + a)/((d^2*tan(b*x + a)^2 + d^2) 
*b) - 42*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2 
*sqrt(d*tan(b*x + a)))/sqrt(abs(d)))/(b*d) - 42*sqrt(2)*abs(d)^(3/2)*arcta 
n(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d) 
))/(b*d) + 21*sqrt(2)*abs(d)^(3/2)*log(d*tan(b*x + a) + sqrt(2)*sqrt(d*tan 
(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d) - 21*sqrt(2)*abs(d)^(3/2)*log(d*ta 
n(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d) + 3 
2*sqrt(d*tan(b*x + a))*tan(b*x + a)/b)*d^2
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(d)*int(sqrt(tan(a + b*x))*sin(a + b*x)**2*tan(a + b*x)**2,x)*d**2