\(\int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [64]

Optimal result

Integrand size = 21, antiderivative size = 224 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {45 d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}+\sqrt {d} \tan (a+b x)}\right )}{32 \sqrt {2} b}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3} \] Output:

45/64*d^(3/2)*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b-45/ 
64*d^(3/2)*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))*2^(1/2)/b-45/64* 
d^(3/2)*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+45/16*d*(d*tan(b*x+a))^(1/2)/b-9/16*cos(b*x+a)^2*(d*tan(b*x+a)) 
^(5/2)/b/d-1/4*cos(b*x+a)^4*(d*tan(b*x+a))^(9/2)/b/d^3
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.55 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {d \csc (a+b x) \left (-143 \sin (a+b x)-45 \arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}+45 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}-14 \sin (3 (a+b x))+\sin (5 (a+b x))\right ) \sqrt {d \tan (a+b x)}}{64 b} \] Input:

Integrate[Sin[a + b*x]^4*(d*Tan[a + b*x])^(3/2),x]
 

Output:

-1/64*(d*Csc[a + b*x]*(-143*Sin[a + b*x] - 45*ArcSin[Cos[a + b*x] - Sin[a 
+ b*x]]*Sqrt[Sin[2*(a + b*x)]] + 45*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt 
[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]] - 14*Sin[3*(a + b*x)] + Sin[5*( 
a + b*x)])*Sqrt[d*Tan[a + b*x]])/b
 

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.24, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3071, 252, 252, 262, 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.

\(\displaystyle \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (a+b x)^4 (d \tan (a+b x))^{3/2}dx\)

\(\Big \downarrow \) 3071

\(\displaystyle \frac {d \int \frac {(d \tan (a+b x))^{11/2}}{\left (\tan ^2(a+b x) d^2+d^2\right )^3}d(d \tan (a+b x))}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {d \left (\frac {9}{8} \int \frac {(d \tan (a+b x))^{7/2}}{\left (\tan ^2(a+b x) d^2+d^2\right )^2}d(d \tan (a+b x))-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \int \frac {(d \tan (a+b x))^{3/2}}{\tan ^2(a+b x) d^2+d^2}d(d \tan (a+b x))-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-d^2 \int \frac {1}{\sqrt {d \tan (a+b x)} \left (\tan ^2(a+b x) d^2+d^2\right )}d(d \tan (a+b x))\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \int \frac {1}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(a+b x)+d}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(a+b x)-1}d\left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {\int \frac {d-d^2 \tan ^2(a+b x)}{d^4 \tan ^4(a+b x)+d^2}d\sqrt {d \tan (a+b x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)-\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}}{d^2 \tan ^2(a+b x)+\sqrt {2} d^{3/2} \tan (a+b x)+d}d\sqrt {d \tan (a+b x)}}{2 \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {d \left (\frac {9}{8} \left (\frac {5}{4} \left (2 \sqrt {d \tan (a+b x)}-2 d^2 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (a+b x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (a+b x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (a+b x)+d^2 \tan ^2(a+b x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )\right )-\frac {(d \tan (a+b x))^{5/2}}{2 \left (d^2 \tan ^2(a+b x)+d^2\right )}\right )-\frac {(d \tan (a+b x))^{9/2}}{4 \left (d^2 \tan ^2(a+b x)+d^2\right )^2}\right )}{b}\)

Input:

Int[Sin[a + b*x]^4*(d*Tan[a + b*x])^(3/2),x]
 

Output:

(d*(-1/4*(d*Tan[a + b*x])^(9/2)/(d^2 + d^2*Tan[a + b*x]^2)^2 + (9*(-1/2*(d 
*Tan[a + b*x])^(5/2)/(d^2 + d^2*Tan[a + b*x]^2) + (5*(-2*d^2*((-(ArcTan[1 
- Sqrt[2]*Sqrt[d]*Tan[a + b*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*Sq 
rt[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 
*Sqrt[d*Tan[a + b*x]]))/4))/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 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 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] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1374\) vs. \(2(174)=348\).

Time = 11.26 (sec) , antiderivative size = 1375, normalized size of antiderivative = 6.14

method	result	size
default	\(\text {Expression too large to display}\)	\(1375\)

Input:

int(sin(b*x+a)^4*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/256/b*(d*tan(b*x+a))^(1/2)*d*(2^(1/2)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x 
+a)+1)^2)^(1/2)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)-2*sin(b*x+a)*(-2*s 
in(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)))*(-180*cos(b*x+a)^2-180*cos(b*x+a)-45*csc(b*x+a)* 
(4*cos(b*x+a)^4-4*cos(b*x+a)^2-1)*(cos(b*x+a)+1))+csc(b*x+a)*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)))*(-s 
in(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*(360*cos(b*x+a)^5+360*cos(b*x 
+a)^4-360*cos(b*x+a)^3+360*cos(b*x+a)^2*sin(b*x+a)-360*cos(b*x+a)^2+360*si 
n(b*x+a)*cos(b*x+a))+2^(1/2)*(-2*sin(b*x+a)*cos(b*x+a)/(cos(b*x+a)+1)^2)^( 
1/2)*ln(-(cot(b*x+a)*cos(b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-2*sin(b*x+a)*c 
os(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)))*(180*cos(b*x+a)^2+180*cos(b*x+a)+45*csc(b*x+a)*(4*cos(b*x+a 
)^4-4*cos(b*x+a)^2-1)*(cos(b*x+a)+1))+csc(b*x+a)*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)*co 
s(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*(-360*cos(b*x+a)^5-360*cos(b*x+a)^4+360*c 
os(b*x+a)^3-360*cos(b*x+a)^2*sin(b*x+a)+360*cos(b*x+a)^2-360*sin(b*x+a)*co 
s(b*x+a))+2^(1/2)*(-2*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+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 451 vs. \(2 (174) = 348\).

Time = 0.22 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.01 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {90 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )}}{2 \, \sqrt {d} \sin \left (b x + a\right )}\right ) - 45 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {2 \, d \cos \left (b x + a\right )^{2} - 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \sqrt {2} \sqrt {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} - 2 \, d}{2 \, {\left (d \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - d\right )}}\right ) - 45 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (-\frac {2 \, d \cos \left (b x + a\right )^{2} - 2 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - \sqrt {2} \sqrt {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} - 2 \, d}{2 \, {\left (d \cos \left (b x + a\right )^{2} + d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - d\right )}}\right ) + 45 \, \sqrt {2} d^{\frac {3}{2}} \log \left (4 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} + d\right ) - 45 \, \sqrt {2} d^{\frac {3}{2}} \log \left (4 \, d \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, \sqrt {2} {\left (\cos \left (b x + a\right )^{2} + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )} \sqrt {d} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} + d\right ) + 16 \, {\left (4 \, d \cos \left (b x + a\right )^{4} - 17 \, d \cos \left (b x + a\right )^{2} - 32 \, d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{256 \, b} \] Input:

integrate(sin(b*x+a)^4*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

-1/256*(90*sqrt(2)*d^(3/2)*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))) - 45*sqrt(2)* 
d^(3/2)*arctan(1/2*(2*d*cos(b*x + a)^2 - 2*d*cos(b*x + a)*sin(b*x + a) + s 
qrt(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)) - 45*sqrt(2)*d^(3/2)*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*si 
n(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)) + 45*sqrt(2)*d^(3/2)*log(4*d*cos(b*x + a)*sin(b*x + a) + 2*sqr 
t(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) - 45*sqrt(2)*d^(3/2)*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*(4*d*cos(b*x + a)^4 - 17*d*cos(b*x 
+ a)^2 - 32*d)*sqrt(d*sin(b*x + a)/cos(b*x + a)))/b
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(sin(b*x+a)**4*(d*tan(b*x+a))**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.05 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {90 \, \sqrt {2} d^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 90 \, \sqrt {2} d^{\frac {13}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 45 \, \sqrt {2} d^{\frac {13}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 45 \, \sqrt {2} d^{\frac {13}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 256 \, \sqrt {d \tan \left (b x + a\right )} d^{6} - \frac {8 \, {\left (17 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{8} + 13 \, \sqrt {d \tan \left (b x + a\right )} d^{10}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{128 \, b d^{5}} \] Input:

integrate(sin(b*x+a)^4*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

-1/128*(90*sqrt(2)*d^(13/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d 
*tan(b*x + a)))/sqrt(d)) + 90*sqrt(2)*d^(13/2)*arctan(-1/2*sqrt(2)*(sqrt(2 
)*sqrt(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 45*sqrt(2)*d^(13/2)*log(d*t 
an(b*x + a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - 45*sqrt(2)*d^(13 
/2)*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - 256*s 
qrt(d*tan(b*x + a))*d^6 - 8*(17*(d*tan(b*x + a))^(5/2)*d^8 + 13*sqrt(d*tan 
(b*x + a))*d^10)/(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 = 0.16 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.20 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {\frac {90 \, \sqrt {2} d^{2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {90 \, \sqrt {2} d^{2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {45 \, \sqrt {2} d^{2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {45 \, \sqrt {2} d^{2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {256 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{b} - \frac {8 \, {\left (17 \, \sqrt {d \tan \left (b x + a\right )} d^{6} \tan \left (b x + a\right )^{2} + 13 \, \sqrt {d \tan \left (b x + a\right )} d^{6}\right )}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b}}{128 \, d} \] Input:

integrate(sin(b*x+a)^4*(d*tan(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

-1/128*(90*sqrt(2)*d^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 + 90*sqrt(2)*d^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 + 45*sqrt(2)*d^2*sqrt(abs(d))*log(d*tan(b*x + a) + sqrt(2)*sqrt 
(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/b - 45*sqrt(2)*d^2*sqrt(abs(d))*lo 
g(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/b - 
 256*sqrt(d*tan(b*x + a))*d^2/b - 8*(17*sqrt(d*tan(b*x + a))*d^6*tan(b*x + 
 a)^2 + 13*sqrt(d*tan(b*x + a))*d^6)/((d^2*tan(b*x + a)^2 + d^2)^2*b))/d
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=\int {\sin \left (a+b\,x\right )}^4\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2} \,d x \] Input:

int(sin(a + b*x)^4*(d*tan(a + b*x))^(3/2),x)
 

Output:

int(sin(a + b*x)^4*(d*tan(a + b*x))^(3/2), x)
 

Reduce [F]

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

int(sin(b*x+a)^4*(d*tan(b*x+a))^(3/2),x)
 

Output:

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