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

Optimal. Leaf size=277 \[ \frac {45 d^{3/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}+\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}-\frac {45 d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{64 \sqrt {2} b}+\frac {45 d \sqrt {d \tan (a+b x)}}{16 b}-\frac {9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac {\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3} \]

[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

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Rubi [A]
time = 0.14, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {2671, 294, 327, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {45 d^{3/2} \text {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 {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} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\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} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{32 \sqrt {2} b}-\frac {45 d^{3/2} \tan ^{-1}\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*}

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Mathematica [A]
time = 0.82, size = 123, normalized size = 0.44 \begin {gather*} -\frac {d \csc (a+b x) \left (-143 \sin (a+b x)-45 \text {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} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 3.80, size = 702, normalized size = 2.53

method	result	size
default	\(\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (45 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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-45 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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-8 \sqrt {2}\, \left (\cos ^{5}\left (b x +a \right )\right )-90 \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 )}}\, \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 )+45 \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 {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+45 \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 {\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}+34 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-34 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+64 \cos \left (b x +a \right ) \sqrt {2}-64 \sqrt {2}\right ) \cos \left (b x +a \right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{64 b \sin \left (b x +a \right )^{5}}\)	\(702\)

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))*(45*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))*si
n(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)*((cos(b*x+a)-1+sin(b*
x+a))/sin(b*x+a))^(1/2)-45*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))*si
n(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)*((cos(b*x+a)-1+sin(b*
x+a))/sin(b*x+a))^(1/2)-8*2^(1/2)*cos(b*x+a)^5-90*((-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)*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))+45*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)*((cos(b*x+a)-
1+sin(b*x+a))/sin(b*x+a))^(1/2)+45*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)*((cos(b*x+a)-1+
sin(b*x+a))/sin(b*x+a))^(1/2)+8*cos(b*x+a)^4*2^(1/2)+34*cos(b*x+a)^3*2^(1/2)-34*cos(b*x+a)^2*2^(1/2)+64*cos(b*
x+a)*2^(1/2)-64*2^(1/2))*cos(b*x+a)*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/sin(b*x+a)^5*2^(1/2)

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Maxima [A]
time = 0.49, size = 235, normalized size = 0.85 \begin {gather*} -\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}} \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*(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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1580 vs. \(2 (213) = 426\).
time = 42.87, size = 1580, normalized size = 5.70 \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]

1/256*(90*sqrt(2)*(d^6/b^4)^(1/4)*b*arctan(1/2*(2*d^10*sin(b*x + a) + sqrt(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a
)*sin(b*x + a) + d^10 + 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b
^3*d^5*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^3*cos(b*x + a) + sqrt(2
)*(d^6/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d^7*cos(b*x + a)^3 - b^2*d^7*co
s(b*x + a))*sqrt(d^6/b^4) + (sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(
b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/((2*d^10*cos(b*x + a)^2 - d^10)*sin(b*x + a))) + 90*sqrt(2)*(d^6/
b^4)^(1/4)*b*arctan(-1/2*(2*d^10*sin(b*x + a) - sqrt(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d^10
- 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)
*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^3*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*
sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d^7*cos(b*x + a)^3 - b^2*d^7*cos(b*x + a))*sqrt(d^6/b
^4) - (sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(b*x + a))*sqrt(d*sin(b
*x + a)/cos(b*x + a)))/((2*d^10*cos(b*x + a)^2 - d^10)*sin(b*x + a))) - 90*sqrt(2)*(d^6/b^4)^(1/4)*b*arctan(1/
2*(sqrt(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d^10 - 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x +
a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(2*d^5*si
n(b*x + a) + (sqrt(2)*(d^6/b^4)^(1/4)*b*d^3*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*si
n(b*x + a)/cos(b*x + a))) - (sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a) - sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(
b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(d^10*sin(b*x + a))) - 90*sqrt(2)*(d^6/b^4)^(1/4)*b*arctan(-1/2*(
sqrt(4*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + d^10 + 2*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*
sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(2*d^5*sin(b
*x + a) - (sqrt(2)*(d^6/b^4)^(1/4)*b*d^3*cos(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*sin(b*x + a))*sqrt(d*sin(b
*x + a)/cos(b*x + a))) + (sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a) - sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*sin(b*x
+ a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(d^10*sin(b*x + a))) - 45*sqrt(2)*(d^6/b^4)^(1/4)*b*log(33215062500*
sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + 8303765625*d^10 + 16607531250*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8
*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))
) + 45*sqrt(2)*(d^6/b^4)^(1/4)*b*log(33215062500*sqrt(d^6/b^4)*b^2*d^7*cos(b*x + a)*sin(b*x + a) + 8303765625*
d^10 - 16607531250*(sqrt(2)*(d^6/b^4)^(1/4)*b*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^6/b^4)^(3/4)*b^3*d^5*
cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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]

Exception raised: SystemError >> excessive stack use: stack is 4370 deep

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Giac [A]
time = 0.53, size = 252, normalized size = 0.91 \begin {gather*} -\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 )} \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*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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\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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