3.1.94 \(\int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) [94]

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

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2671, 294, 296, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {3 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b d^{3/2}}+\frac {3 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{32 \sqrt {2} b d^{3/2}}+\frac {3 \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b d^{3/2}}-\frac {3 \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{64 \sqrt {2} b d^{3/2}}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{3/2}}{4 b d^3}+\frac {3 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b*d^(3/2)) + (3*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b*d^(3/2)) + (3*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*Sqrt[d*Tan[
a + b*x]]])/(64*Sqrt[2]*b*d^(3/2)) - (3*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a + b*x]]])/(6
4*Sqrt[2]*b*d^(3/2)) + (3*Cos[a + b*x]^2*(d*Tan[a + b*x])^(3/2))/(16*b*d^3) - (Cos[a + b*x]^4*(d*Tan[a + b*x])
^(3/2))/(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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 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*} \int \frac {\sin ^4(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=\frac {d \text {Subst}\left (\int \frac {x^{5/2}}{\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))^{3/2}}{4 b d^3}+\frac {(3 d) \text {Subst}\left (\int \frac {\sqrt {x}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{8 b}\\ &=\frac {3 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d^3}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{3/2}}{4 b d^3}+\frac {3 \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{32 b d}\\ &=\frac {3 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d^3}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{3/2}}{4 b d^3}+\frac {3 \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{16 b d}\\ &=\frac {3 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d^3}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{3/2}}{4 b d^3}-\frac {3 \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b d}+\frac {3 \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{32 b d}\\ &=\frac {3 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d^3}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{3/2}}{4 b d^3}+\frac {3 \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 d^{3/2}}+\frac {3 \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 d^{3/2}}+\frac {3 \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 d}+\frac {3 \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 d}\\ &=\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b d^{3/2}}-\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b d^{3/2}}+\frac {3 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d^3}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{3/2}}{4 b d^3}+\frac {3 \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 d^{3/2}}-\frac {3 \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 d^{3/2}}\\ &=-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b d^{3/2}}+\frac {3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b d^{3/2}}+\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b d^{3/2}}-\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b d^{3/2}}+\frac {3 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d^3}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{3/2}}{4 b d^3}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 123, normalized size = 0.48 \begin {gather*} \frac {\csc (a+b x) \left (\cos (a+b x)-2 \cos (3 (a+b x))+\cos (5 (a+b x))-3 \text {ArcSin}(\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}-3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}\right ) \sqrt {d \tan (a+b x)}}{64 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x]*(Cos[a + b*x] - 2*Cos[3*(a + b*x)] + Cos[5*(a + b*x)] - 3*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Sq
rt[Sin[2*(a + b*x)]] - 3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]])*Sqr
t[d*Tan[a + b*x]])/(64*b*d^2)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.32, size = 550, normalized size = 2.14

method	result	size
default	\(\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (3 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 ) \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-3 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 ) \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-8 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}+3 \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 ) \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+3 \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 ) \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+8 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+6 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-6 \cos \left (b x +a \right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{64 b \cos \left (b x +a \right )^{2} \sin \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}}}\)	\(550\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/b*(-1+cos(b*x+a))*(3*I*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))*((-
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)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*
x+a))^(1/2)-3*I*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))*((-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)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-
8*cos(b*x+a)^4*2^(1/2)+3*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))*((-1+c
os(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a
))^(1/2)+3*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))*((-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)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)+8*cos
(b*x+a)^3*2^(1/2)+6*cos(b*x+a)^2*2^(1/2)-6*cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2/cos(b*x+a)^2/sin(b*x+a)/(d*sin
(b*x+a)/cos(b*x+a))^(3/2)*2^(1/2)

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Maxima [A]
time = 0.49, size = 225, normalized size = 0.88 \begin {gather*} \frac {3 \, d^{4} {\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 {8 \, {\left (3 \, \left (d \tan \left (b x + a\right )\right )^{\frac {7}{2}} d^{4} - \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{6}\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(3*d^4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) + 2*sqr
t(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)) + 8*(3*(d*tan(b*x + a))^(7/2)*d^4 - (d*tan(b*x + a))^(3/2)*d^6)/(d^4*tan(b*x + a)^
4 + 2*d^4*tan(b*x + a)^2 + d^4))/(b*d^5)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1871 vs. \(2 (197) = 394\).
time = 62.50, size = 1871, normalized size = 7.28 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/512*(12*sqrt(2)*b*d^2*(1/(b^4*d^6))^(1/4)*arctan((sqrt(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a)
- 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x
+ a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*(b^2*d^3*sqrt(1/(b^4*d^6)) + 2*cos(b*x + a)*sin(b*x + a) + (sqr
t(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)*sin(b*x + a))*s
qrt(d*sin(b*x + a)/cos(b*x + a))) + (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b
*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(2*cos(b*x + a)^2 - 1)) + 12*sqrt(2)
*b*d^2*(1/(b^4*d^6))^(1/4)*arctan(-(sqrt(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^
3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*s
in(b*x + a)/cos(b*x + a)) + 1)*(b^2*d^3*sqrt(1/(b^4*d^6)) + 2*cos(b*x + a)*sin(b*x + a) - (sqrt(2)*b^3*d^4*(1/
(b^4*d^6))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x +
a)/cos(b*x + a))) - (sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))
^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(2*cos(b*x + a)^2 - 1)) + 12*sqrt(2)*b*d^2*(1/(b^4*d
^6))^(1/4)*arctan(1/2*((sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*sin
(b*x + a))*sqrt(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)
*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*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^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(
b^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 4*(b^2*d^3*cos(b*x + a)^3 - b^2*d^3*cos(b*x
+ a))*sqrt(1/(b^4*d^6)) - 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) + 12*sqrt(2)*b*d^2*(1/(b^4*d^
6))^(1/4)*arctan(1/2*((sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*sin(
b*x + a))*sqrt(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*
cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*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^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a) + sqrt(2)*b*d*(1/(b
^4*d^6))^(1/4)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d^3*cos(b*x + a)^3 - b^2*d^3*cos(b*x +
a))*sqrt(1/(b^4*d^6)) + 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) - 3*sqrt(2)*b*d^2*(1/(b^4*d^6)
)^(1/4)*log(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos
(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1
) + 3*sqrt(2)*b*d^2*(1/(b^4*d^6))^(1/4)*log(4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)
*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(
d*sin(b*x + a)/cos(b*x + a)) + 1) - 3*sqrt(2)*b*d^2*(1/(b^4*d^6))^(1/4)*log(1/4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(
b*x + a)*sin(b*x + a) + 1/8*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b
^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1/16) + 3*sqrt(2)*b*d^2*(1/(b^4*d^6))^(1/4)
*log(1/4*b^2*d^3*sqrt(1/(b^4*d^6))*cos(b*x + a)*sin(b*x + a) - 1/8*(sqrt(2)*b^3*d^4*(1/(b^4*d^6))^(3/4)*cos(b*
x + a)*sin(b*x + a) + sqrt(2)*b*d*(1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1/16
) - 32*(4*cos(b*x + a)^3 - 3*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))*sin(b*x + a))/(b*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{4}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.76, size = 257, normalized size = 1.00 \begin {gather*} \frac {\frac {6 \, \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^{2}} + \frac {6 \, \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^{2}} - \frac {3 \, \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^{2}} + \frac {3 \, \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^{2}} + \frac {8 \, {\left (3 \, \sqrt {d \tan \left (b x + a\right )} d^{3} \tan \left (b x + a\right )^{3} - \sqrt {d \tan \left (b x + a\right )} d^{3} \tan \left (b x + a\right )\right )}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )}^{2} b}}{128 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(6*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^2) + 6*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^2) - 3*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^2) + 3*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^2) + 8*(3*sqrt(d*tan(b*x + a))*d^3*tan(b*x + a)^3 - sqrt(d*tan(b*x + a))*d^3*tan(b*x + a))/((d^2*tan(b*x +
a)^2 + d^2)^2*b))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^4}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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