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

Optimal result

Integrand size = 21, antiderivative size = 277 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {45 d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}+\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3} \]

[Out]

45/64*d^(3/2)*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)-45/64*d^(3/2)*arctan(1+2^(1/2)*(d*tan(b
*x+a))^(1/2)/d^(1/2))/b*2^(1/2)+45/128*d^(3/2)*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)-45/128*d^(3/2)*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)+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

Rubi [A] (verified)

Time = 0.23 (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, 217, 1179, 642, 1176, 631, 210} \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=\frac {45 d^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{32 \sqrt {2} b}+\frac {45 d^{3/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d} \]

[In]

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

[Out]

(45*d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) - (45*d^(3/2)*ArcTan[1 + (Sqrt[
2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) + (45*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*S
qrt[d*Tan[a + b*x]]])/(64*Sqrt[2]*b) - (45*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a +
b*x]]])/(64*Sqrt[2]*b) + (45*d*Sqrt[d*Tan[a + b*x]])/(16*b) - (9*Cos[a + b*x]^2*(d*Tan[a + b*x])^(5/2))/(16*b
*d) - (Cos[a + b*x]^4*(d*Tan[a + b*x])^(9/2))/(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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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 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 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^{11/2}}{\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))^{9/2}}{4 b d^3}+\frac {(9 d) \text {Subst}\left (\int \frac {x^{7/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{8 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}+\frac {(45 d) \text {Subst}\left (\int \frac {x^{3/2}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{32 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}-\frac {\left (45 d^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (a+b x)\right )}{32 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}-\frac {\left (45 d^3\right ) \text {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{16 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}-\frac {\left (45 d^2\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 (45 d^2\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 {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}+\frac {\left (45 d^{3/2}\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 (45 d^{3/2}\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 (45 d^2\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 (45 d^2\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 {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {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}-\frac {\left (45 d^{3/2}\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 (45 d^{3/2}\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 {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} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.44 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {d \csc (a+b x) \left (-143 \sin (a+b x)-45 \arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}+45 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}-14 \sin (3 (a+b x))+\sin (5 (a+b x))\right ) \sqrt {d \tan (a+b x)}}{64 b} \]

[In]

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

[Out]

-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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(889\) vs. \(2(213)=426\).

Time = 3.46 (sec) , antiderivative size = 890, normalized size of antiderivative = 3.21

method	result	size
default	\(\text {Expression too large to display}\)	\(890\)

[In]

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

[Out]

1/128/b*(d*tan(b*x+a))^(1/2)*(-16*2^(1/2)*cos(b*x+a)^4+68*cos(b*x+a)^2*2^(1/2)+45*cot(b*x+a)*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)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-45*cot(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)-90*cot(b*x+a)*(-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)))+90*cot(b*x+a)*(-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)))+45*csc(b*x+a)*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)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-45*csc(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)-90*csc(b*x+a)*
(-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)))+90*csc(b*x+a)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arct
an((-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))))*d*2^(1
/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 954, normalized size of antiderivative = 3.44 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/256*(45*I*(-d^6/b^4)^(1/4)*b*log(182250*d^5*cos(b*x + a)^2 + 182250*sqrt(-d^6/b^4)*b^2*d^2*cos(b*x + a)*sin
(b*x + a) - 91125*d^5 - 182250*(I*(-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) - I*(-d^6/b^4)^(3/4)*b^3*co
s(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 45*I*(-d^6/b^4)^(1/4)*b*log(182250*d^5*cos(b*x + a)^2 + 182
250*sqrt(-d^6/b^4)*b^2*d^2*cos(b*x + a)*sin(b*x + a) - 91125*d^5 - 182250*(-I*(-d^6/b^4)^(1/4)*b*d^3*cos(b*x +
a)*sin(b*x + a) + I*(-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 45*(-d^6/b^4)^(
1/4)*b*log(182250*d^5*cos(b*x + a)^2 - 182250*sqrt(-d^6/b^4)*b^2*d^2*cos(b*x + a)*sin(b*x + a) - 91125*d^5 + 1
82250*((-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) + (-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x
+ a)/cos(b*x + a))) + 45*(-d^6/b^4)^(1/4)*b*log(182250*d^5*cos(b*x + a)^2 - 182250*sqrt(-d^6/b^4)*b^2*d^2*cos(
b*x + a)*sin(b*x + a) - 91125*d^5 - 182250*((-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) + (-d^6/b^4)^(3/4
)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 45*(-d^6/b^4)^(1/4)*b*log(-91125*d^5 + 182250*((-d^
6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) - (-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*
x + a))) + 45*(-d^6/b^4)^(1/4)*b*log(-91125*d^5 - 182250*((-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) - (
-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) + 45*I*(-d^6/b^4)^(1/4)*b*log(-91125*d^
5 - 182250*(I*(-d^6/b^4)^(1/4)*b*d^3*cos(b*x + a)*sin(b*x + a) + I*(-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d
*sin(b*x + a)/cos(b*x + a))) - 45*I*(-d^6/b^4)^(1/4)*b*log(-91125*d^5 - 182250*(-I*(-d^6/b^4)^(1/4)*b*d^3*cos(
b*x + a)*sin(b*x + a) - I*(-d^6/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) + 16*(4*d*co
s(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. \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.85 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {90 \, \sqrt {2} d^{\frac {13}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 90 \, \sqrt {2} d^{\frac {13}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 45 \, \sqrt {2} d^{\frac {13}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 45 \, \sqrt {2} d^{\frac {13}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 256 \, \sqrt {d \tan \left (b x + a\right )} d^{6} - \frac {8 \, {\left (17 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{8} + 13 \, \sqrt {d \tan \left (b x + a\right )} d^{10}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{128 \, b d^{5}} \]

[In]

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

[Out]

-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
*tan(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)*s
qrt(d*tan(b*x + a))*sqrt(d) + d) - 256*sqrt(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)

none

Time = 0.38 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.91 \[ \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx=-\frac {1}{128} \, d {\left (\frac {90 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {90 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b} + \frac {45 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {45 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b} - \frac {256 \, \sqrt {d \tan \left (b x + a\right )}}{b} - \frac {8 \, {\left (17 \, \sqrt {d \tan \left (b x + a\right )} d^{4} \tan \left (b x + a\right )^{2} + 13 \, \sqrt {d \tan \left (b x + a\right )} d^{4}\right )}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b}\right )} \]

[In]

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

[Out]

-1/128*d*(90*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 + 90*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 + 45*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 -
45*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 - 256*sqrt(
d*tan(b*x + a))/b - 8*(17*sqrt(d*tan(b*x + a))*d^4*tan(b*x + a)^2 + 13*sqrt(d*tan(b*x + a))*d^4)/((d^2*tan(b*x
+ a)^2 + d^2)^2*b))

Mupad [F(-1)]

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

[In]

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

[Out]

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