\(\int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx\) [73]

Optimal result

Integrand size = 21, antiderivative size = 277 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {77 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {77 d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {77 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {77 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3} \]

[Out]

77/64*d^(5/2)*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)-77/64*d^(5/2)*arctan(1+2^(1/2)*(d*tan(b
*x+a))^(1/2)/d^(1/2))/b*2^(1/2)-77/128*d^(5/2)*ln(d^(1/2)-2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2
^(1/2)+77/128*d^(5/2)*ln(d^(1/2)+2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2^(1/2)+77/48*d*(d*tan(b*x
+a))^(3/2)/b-11/16*cos(b*x+a)^2*(d*tan(b*x+a))^(7/2)/b/d-1/4*cos(b*x+a)^4*(d*tan(b*x+a))^(11/2)/b/d^3

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2671, 294, 327, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {77 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {77 d^{5/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{32 \sqrt {2} b}-\frac {77 d^{5/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}+\frac {77 d^{5/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}+\frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d} \]

[In]

Int[Sin[a + b*x]^4*(d*Tan[a + b*x])^(5/2),x]

[Out]

(77*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) - (77*d^(5/2)*ArcTan[1 + (Sqrt[
2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) - (77*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*S
qrt[d*Tan[a + b*x]]])/(64*Sqrt[2]*b) + (77*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a +
b*x]]])/(64*Sqrt[2]*b) + (77*d*(d*Tan[a + b*x])^(3/2))/(48*b) - (11*Cos[a + b*x]^2*(d*Tan[a + b*x])^(7/2))/(1
6*b*d) - (Cos[a + b*x]^4*(d*Tan[a + b*x])^(11/2))/(4*b*d^3)

Rule 210

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 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {d \text {Subst}\left (\int \frac {x^{13/2}}{\left (d^2+x^2\right )^3} \, dx,x,d \tan (a+b x)\right )}{b} \\ & = -\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}+\frac {(11 d) \text {Subst}\left (\int \frac {x^{9/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{8 b} \\ & = -\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}+\frac {(77 d) \text {Subst}\left (\int \frac {x^{5/2}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{32 b} \\ & = \frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}-\frac {\left (77 d^3\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{32 b} \\ & = \frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}-\frac {\left (77 d^3\right ) \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{16 b} \\ & = \frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}+\frac {\left (77 d^3\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b}-\frac {\left (77 d^3\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b} \\ & = \frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}-\frac {\left (77 d^{5/2}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {\left (77 d^{5/2}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {\left (77 d^3\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 b}-\frac {\left (77 d^3\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{64 b} \\ & = -\frac {77 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {77 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3}-\frac {\left (77 d^{5/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}+\frac {\left (77 d^{5/2}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b} \\ & = \frac {77 d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {77 d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {77 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {77 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {77 d (d \tan (a+b x))^{3/2}}{48 b}-\frac {11 \cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{11/2}}{4 b d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.51 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {d \left (128+204 \cos ^2(a+b x)+231 \arcsin (\cos (a+b x)-\sin (a+b x)) \cot (a+b x) \csc (a+b x) \sqrt {\sin (2 (a+b x))}+231 \cot (a+b x) \csc (a+b x) \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}-6 \cot (a+b x) \sin (4 (a+b x))\right ) (d \tan (a+b x))^{3/2}}{192 b} \]

[In]

Integrate[Sin[a + b*x]^4*(d*Tan[a + b*x])^(5/2),x]

[Out]

(d*(128 + 204*Cos[a + b*x]^2 + 231*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Cot[a + b*x]*Csc[a + b*x]*Sqrt[Sin[2*(a
+ b*x)]] + 231*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)]] - 6*Cot[a + b*x]*Sin[4*(a + b*x)])*(d*Tan[a + b*x])^(3/2))/(192*b)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(525\) vs. \(2(213)=426\).

Time = 48.45 (sec) , antiderivative size = 526, normalized size of antiderivative = 1.90

method	result	size
default	\(-\frac {\tan \left (b x +a \right ) \sqrt {d \tan \left (b x +a \right )}\, \left (48 \left (\cos ^{5}\left (b x +a \right )\right ) \sqrt {2}-48 \sqrt {2}\, \left (\cos ^{4}\left (b x +a \right )\right )-228 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+228 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-462 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \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 {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\cos \left (b x +a \right )-1}{-1+\cos \left (b x +a \right )}\right ) \cos \left (b x +a \right )+462 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \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 {2}\, \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}+\cos \left (b x +a \right )-1}{-1+\cos \left (b x +a \right )}\right ) \cos \left (b x +a \right )-231 \cos \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \ln \left (2 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \csc \left (b x +a \right )+2 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \cot \left (b x +a \right )-2 \cot \left (b x +a \right )+2\right )+231 \cos \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \ln \left (-2 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \csc \left (b x +a \right )-2 \sqrt {-\frac {\cos \left (b x +a \right ) \sin \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \sqrt {2}\, \cot \left (b x +a \right )-2 \cot \left (b x +a \right )+2\right )-128 \sqrt {2}\, \cos \left (b x +a \right )+128 \sqrt {2}\right ) d^{2} \sqrt {2}}{384 b \left (-1+\cos \left (b x +a \right )\right )}\)	\(526\)

[In]

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

[Out]

-1/384/b*tan(b*x+a)*(d*tan(b*x+a))^(1/2)*(48*cos(b*x+a)^5*2^(1/2)-48*2^(1/2)*cos(b*x+a)^4-228*cos(b*x+a)^3*2^(
1/2)+228*cos(b*x+a)^2*2^(1/2)-462*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((-sin(b*x+a)*2^(1/2)*
(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*x+a)+462*(-cos(b*x+a)*sin
(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos
(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*x+a)-231*cos(b*x+a)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(2*(-c
os(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*2^(1/2)*csc(b*x+a)+2*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1
/2)*2^(1/2)*cot(b*x+a)-2*cot(b*x+a)+2)+231*cos(b*x+a)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-2*(-
cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*2^(1/2)*csc(b*x+a)-2*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(
1/2)*2^(1/2)*cot(b*x+a)-2*cot(b*x+a)+2)-128*2^(1/2)*cos(b*x+a)+128*2^(1/2))/(-1+cos(b*x+a))*d^2*2^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 1048, normalized size of antiderivative = 3.78 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/768*(231*(-d^10/b^4)^(1/4)*b*cos(b*x + a)*log(-456533/2*d^8*cos(b*x + a)*sin(b*x + a) + 456533/4*(2*b^2*d^3
*cos(b*x + a)^2 - b^2*d^3)*sqrt(-d^10/b^4) + 456533/2*((-d^10/b^4)^(1/4)*b*d^5*cos(b*x + a)*sin(b*x + a) - (-d
^10/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 231*(-d^10/b^4)^(1/4)*b*cos(b*x + a)*l
og(-456533/2*d^8*cos(b*x + a)*sin(b*x + a) + 456533/4*(2*b^2*d^3*cos(b*x + a)^2 - b^2*d^3)*sqrt(-d^10/b^4) - 4
56533/2*((-d^10/b^4)^(1/4)*b*d^5*cos(b*x + a)*sin(b*x + a) - (-d^10/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(
b*x + a)/cos(b*x + a))) - 231*I*(-d^10/b^4)^(1/4)*b*cos(b*x + a)*log(-456533/2*d^8*cos(b*x + a)*sin(b*x + a) -
456533/4*(2*b^2*d^3*cos(b*x + a)^2 - b^2*d^3)*sqrt(-d^10/b^4) - 456533/2*(I*(-d^10/b^4)^(1/4)*b*d^5*cos(b*x +
a)*sin(b*x + a) + I*(-d^10/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) + 231*I*(-d^10/b
^4)^(1/4)*b*cos(b*x + a)*log(-456533/2*d^8*cos(b*x + a)*sin(b*x + a) - 456533/4*(2*b^2*d^3*cos(b*x + a)^2 - b^
2*d^3)*sqrt(-d^10/b^4) - 456533/2*(-I*(-d^10/b^4)^(1/4)*b*d^5*cos(b*x + a)*sin(b*x + a) - I*(-d^10/b^4)^(3/4)*
b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 231*(-d^10/b^4)^(1/4)*b*cos(b*x + a)*log(456533*d^8 +
913066*((-d^10/b^4)^(1/4)*b*d^5*cos(b*x + a)^2 - (-d^10/b^4)^(3/4)*b^3*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(
b*x + a)/cos(b*x + a))) + 231*(-d^10/b^4)^(1/4)*b*cos(b*x + a)*log(456533*d^8 - 913066*((-d^10/b^4)^(1/4)*b*d^
5*cos(b*x + a)^2 - (-d^10/b^4)^(3/4)*b^3*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) + 231*I
*(-d^10/b^4)^(1/4)*b*cos(b*x + a)*log(456533*d^8 - 913066*(I*(-d^10/b^4)^(1/4)*b*d^5*cos(b*x + a)^2 + I*(-d^10
/b^4)^(3/4)*b^3*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 231*I*(-d^10/b^4)^(1/4)*b*cos(
b*x + a)*log(456533*d^8 - 913066*(-I*(-d^10/b^4)^(1/4)*b*d^5*cos(b*x + a)^2 - I*(-d^10/b^4)^(3/4)*b^3*cos(b*x
+ a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) + 16*(12*d^2*cos(b*x + a)^4 - 57*d^2*cos(b*x + a)^2 - 32
*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. \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.87 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {231 \, d^{8} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - 256 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{6} - \frac {24 \, {\left (19 \, \left (d \tan \left (b x + a\right )\right )^{\frac {7}{2}} d^{8} + 15 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{10}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{384 \, b d^{5}} \]

[In]

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

[Out]

-1/384*(231*d^8*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) + 2*
sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(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)) - 256*(d*tan(b*x + a))^(3/2)*d^6 - 24*(19*(d*tan(b*x + a))^(7/2)*d^8 + 15*(d*ta
n(b*x + a))^(3/2)*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)

none

Time = 0.35 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.00 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {1}{384} \, d^{2} {\left (\frac {462 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d} + \frac {462 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d} - \frac {231 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d} + \frac {231 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d} - \frac {256 \, \sqrt {d \tan \left (b x + a\right )} \tan \left (b x + a\right )}{b} - \frac {24 \, {\left (19 \, \sqrt {d \tan \left (b x + a\right )} d^{4} \tan \left (b x + a\right )^{3} + 15 \, \sqrt {d \tan \left (b x + a\right )} d^{4} \tan \left (b x + a\right )\right )}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b}\right )} \]

[In]

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

[Out]

-1/384*d^2*(462*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(b*x + a)))/sqrt(a
bs(d)))/(b*d) + 462*sqrt(2)*abs(d)^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/s
qrt(abs(d)))/(b*d) - 231*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) + 231*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) - 256*sqrt(d*tan(b*x + a))*tan(b*x + a)/b - 24*(19*sqrt(d*tan(b*x + a))*d^4*tan(b*x + a)^3 + 15*sqr
t(d*tan(b*x + a))*d^4*tan(b*x + a))/((d^2*tan(b*x + a)^2 + d^2)^2*b))

Mupad [F(-1)]

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

[In]

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

[Out]

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