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

Optimal. Leaf size=257 \[ -\frac {\cos ^4(a+b x) (d \tan (a+b x))^{5/2}}{4 b d^3}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b \sqrt {d}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{32 \sqrt {2} b \sqrt {d}}-\frac {5 \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b \sqrt {d}}+\frac {5 \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b \sqrt {d}}-\frac {5 \cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{16 b d} \]

[Out]

-5/64*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)/d^(1/2)+5/64*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1
/2)/d^(1/2))/b*2^(1/2)/d^(1/2)-5/128*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)/d^(
1/2)+5/128*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)/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*tan(b*x+a))^(5/2)/b/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2591, 288, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\cos ^4(a+b x) (d \tan (a+b x))^{5/2}}{4 b d^3}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b \sqrt {d}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{32 \sqrt {2} b \sqrt {d}}-\frac {5 \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b \sqrt {d}}+\frac {5 \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b \sqrt {d}}-\frac {5 \cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{16 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b*Sqrt[d]) + (5*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b*Sqrt[d]) - (5*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*Sqrt[d*Tan[
a + b*x]]])/(64*Sqrt[2]*b*Sqrt[d]) + (5*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a + b*x]]])/(6
4*Sqrt[2]*b*Sqrt[d]) - (5*Cos[a + b*x]^2*Sqrt[d*Tan[a + b*x]])/(16*b*d) - (Cos[a + b*x]^4*(d*Tan[a + b*x])^(5/
2))/(4*b*d^3)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

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 288

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 329

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 617

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 628

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 1162

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 1165

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 2591

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])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sin ^4(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx &=\frac {d \operatorname {Subst}\left (\int \frac {x^{7/2}}{\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))^{5/2}}{4 b d^3}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {x^{3/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{8 b}\\ &=-\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}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (a+b x)\right )}{32 b}\\ &=-\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}+\frac {(5 d) \operatorname {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{16 b}\\ &=-\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}+\frac {5 \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b}+\frac {5 \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b}\\ &=-\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}+\frac {5 \operatorname {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 {5 \operatorname {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 {5 \operatorname {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 \sqrt {d}}-\frac {5 \operatorname {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 \sqrt {d}}\\ &=-\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b \sqrt {d}}+\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \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}+\frac {5 \operatorname {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 \sqrt {d}}-\frac {5 \operatorname {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 \sqrt {d}}\\ &=-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b \sqrt {d}}+\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b \sqrt {d}}-\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b \sqrt {d}}+\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.71, size = 122, normalized size = 0.47 \[ \frac {\sec (a+b x) \left (-7 \sin (a+b x)-6 \sin (3 (a+b x))+\sin (5 (a+b x))-5 \sqrt {\sin (2 (a+b x))} \sin ^{-1}(\cos (a+b x)-\sin (a+b x))+5 \sqrt {\sin (2 (a+b x))} \log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{64 b \sqrt {d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[a + b*x]*(-7*Sin[a + b*x] - 5*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Sqrt[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]])

________________________________________________________________________________________

fricas [B]  time = 62.82, size = 1456, normalized size = 5.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/256*(10*sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*arctan(1/2*(sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a
) - 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x +
a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b
^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 2*sin(b*x + a)) - (sqrt(2)*b^3*d*(1/(b^4*d^2)
)^(3/4)*sin(b*x + a) - sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/sin(b*x
+ a)) + 10*sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*arctan(1/2*(sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a
) + 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x +
a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b
^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 2*sin(b*x + a)) - (sqrt(2)*b^3*d*(1/(b^4*d^2)
)^(3/4)*sin(b*x + a) - sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/sin(b*x
+ a)) + 10*sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*arctan(1/2*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(
2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b
^3*d*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(
b*x + a)/cos(b*x + a)) + 1)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + (sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a
) + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d*cos(b*x + a)^3 -
b^2*d*cos(b*x + a))*sqrt(1/(b^4*d^2)) + 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) + 10*sqrt(2)*b*
d*(1/(b^4*d^2))^(1/4)*arctan(1/2*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b^4*d^2))^(1
/4)*cos(b*x + a))*sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3
/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)
) + 1)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + (sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b^4
*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 4*(b^2*d*cos(b*x + a)^3 - b^2*d*cos(b*x + a))*s
qrt(1/(b^4*d^2)) - 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) - 5*sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*
log(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2
+ sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1) + 5*sqrt(2)*
b*d*(1/(b^4*d^2))^(1/4)*log(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d*(1/(b^4*d^2
))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x
+ a)) + 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)

________________________________________________________________________________________

giac [A]  time = 1.59, size = 246, normalized size = 0.96 \[ \frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d} + \frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{64 \, b d} + \frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d} - \frac {5 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{128 \, b d} - \frac {9 \, \sqrt {d \tan \left (b x + a\right )} d^{3} \tan \left (b x + a\right )^{2} + 5 \, \sqrt {d \tan \left (b x + a\right )} d^{3}}{16 \, {\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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) + 5/64*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*tan(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)

________________________________________________________________________________________

maple [C]  time = 0.51, size = 688, normalized size = 2.68 \[ \frac {\left (-1+\cos \left (b x +a \right )\right ) \left (5 i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-5 i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+8 \sqrt {2}\, \left (\cos ^{5}\left (b x +a \right )\right )-8 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}-5 \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-5 \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+10 \sin \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-18 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+18 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{64 b \cos \left (b x +a \right ) \sin \left (b x +a \right )^{3} \sqrt {\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/b*(-1+cos(b*x+a))*(5*I*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*(
(1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*sin(b*x+a)*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)
,1/2+1/2*I,1/2*2^(1/2))-5*I*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*(
(1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*sin(b*x+a)*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)
,1/2-1/2*I,1/2*2^(1/2))+8*2^(1/2)*cos(b*x+a)^5-8*cos(b*x+a)^4*2^(1/2)-5*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/
sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(b*x+a)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x
+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)-5*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/si
n(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(b*x+a)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a
))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)+10*sin(b*x+a)*EllipticF(((1-cos(b*x+a)+sin(b
*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a
))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)-18*cos(b*x+a)^3*2^(1/2)+18*cos(b*x+a)^2*2^(1/2))*(cos(b*
x+a)+1)^2/cos(b*x+a)/sin(b*x+a)^3/(d*sin(b*x+a)/cos(b*x+a))^(1/2)*2^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 220, normalized size = 0.86 \[ \frac {10 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 10 \, \sqrt {2} d^{\frac {9}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 5 \, \sqrt {2} d^{\frac {9}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 5 \, \sqrt {2} d^{\frac {9}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \frac {8 \, {\left (9 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{6} + 5 \, \sqrt {d \tan \left (b x + a\right )} d^{8}\right )}}{d^{4} \tan \left (b x + a\right )^{4} + 2 \, d^{4} \tan \left (b x + a\right )^{2} + d^{4}}}{128 \, b d^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(10*sqrt(2)*d^(9/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 10*sqrt(2)*
d^(9/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(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*t
an(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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (a+b\,x\right )}^4}{\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{4}{\left (a + b x \right )}}{\sqrt {d \tan {\left (a + b x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________