3.1.74 \(\int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx\) [74]

Optimal. Leaf size=247 \[ \frac {7 d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}-\frac {7 d^{5/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}-\frac {7 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b}+\frac {7 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b}+\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d} \]

[Out]

7/8*d^(5/2)*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)-7/8*d^(5/2)*arctan(1+2^(1/2)*(d*tan(b*x+a
))^(1/2)/d^(1/2))/b*2^(1/2)-7/16*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)
+7/16*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)+7/6*d*(d*tan(b*x+a))^(3/2)
/b-1/2*cos(b*x+a)^2*(d*tan(b*x+a))^(7/2)/b/d

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Rubi [A]
time = 0.12, antiderivative size = 247, 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, 327, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {7 d^{5/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}-\frac {7 d^{5/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b}-\frac {7 d^{5/2} \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}+\frac {7 d^{5/2} \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b}+\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*d^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(4*Sqrt[2]*b) - (7*d^(5/2)*ArcTan[1 + (Sqrt[2]*
Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(4*Sqrt[2]*b) - (7*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*Sqrt[d
*Tan[a + b*x]]])/(8*Sqrt[2]*b) + (7*d^(5/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a + b*x]]]
)/(8*Sqrt[2]*b) + (7*d*(d*Tan[a + b*x])^(3/2))/(6*b) - (Cos[a + b*x]^2*(d*Tan[a + b*x])^(7/2))/(2*b*d)

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*} \int \sin ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx &=\frac {d \text {Subst}\left (\int \frac {x^{9/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d}+\frac {(7 d) \text {Subst}\left (\int \frac {x^{5/2}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{4 b}\\ &=\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d}-\frac {\left (7 d^3\right ) \text {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{4 b}\\ &=\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d}-\frac {\left (7 d^3\right ) \text {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{2 b}\\ &=\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d}+\frac {\left (7 d^3\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b}-\frac {\left (7 d^3\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b}\\ &=\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d}-\frac {\left (7 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 )}{8 \sqrt {2} b}-\frac {\left (7 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 )}{8 \sqrt {2} b}-\frac {\left (7 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 )}{8 b}-\frac {\left (7 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 )}{8 b}\\ &=-\frac {7 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b}+\frac {7 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b}+\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d}-\frac {\left (7 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 )}{4 \sqrt {2} b}+\frac {\left (7 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 )}{4 \sqrt {2} b}\\ &=\frac {7 d^{5/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}-\frac {7 d^{5/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b}-\frac {7 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b}+\frac {7 d^{5/2} \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b}+\frac {7 d (d \tan (a+b x))^{3/2}}{6 b}-\frac {\cos ^2(a+b x) (d \tan (a+b x))^{7/2}}{2 b d}\\ \end {align*}

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Mathematica [A]
time = 0.42, size = 126, normalized size = 0.51 \begin {gather*} \frac {d \left (16+12 \cos ^2(a+b x)+21 \text {ArcSin}(\cos (a+b x)-\sin (a+b x)) \cot (a+b x) \csc (a+b x) \sqrt {\sin (2 (a+b x))}+21 \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))}\right ) (d \tan (a+b x))^{3/2}}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(16 + 12*Cos[a + b*x]^2 + 21*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Cot[a + b*x]*Csc[a + b*x]*Sqrt[Sin[2*(a +
b*x)]] + 21*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)]])*(d*Tan[a + b*x])^(3/2))/(24*b)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.28, size = 564, normalized size = 2.28

method	result	size
default	\(-\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (21 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 )}}\, \cos \left (b x +a \right )-21 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 )}}\, \cos \left (b x +a \right )+21 \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 )}}\, \cos \left (b x +a \right )+21 \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 )}}\, \cos \left (b x +a \right )-6 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+6 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-8 \cos \left (b x +a \right ) \sqrt {2}+8 \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 {5}{2}} \sqrt {2}}{24 b \sin \left (b x +a \right )^{5}}\)	\(564\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/b*(-1+cos(b*x+a))*(21*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))*c
os(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)-21*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))*c
os(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)+21*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)*cos(b*x+a)+21*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)*cos(b*x+a)-6*cos(b*x+a)^3*2^(1/2)+6*cos(b*x+a)^2*2^(1/2)-8*cos(b*x+a)*2^(1/2)+8*2^(1/2))*cos(b*x+a)*
(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(5/2)/sin(b*x+a)^5*2^(1/2)

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Maxima [A]
time = 0.50, size = 209, normalized size = 0.85 \begin {gather*} -\frac {21 \, d^{6} {\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 {24 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{6}}{d^{2} \tan \left (b x + a\right )^{2} + d^{2}} - 32 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}} d^{4}}{48 \, b d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(21*d^6*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) + 2*sq
rt(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)) - 24*(d*tan(b*x + a))^(3/2)*d^6/(d^2*tan(b*x + a)^2 + d^2) - 32*(d*tan(b*x + a))^
(3/2)*d^4)/(b*d^3)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1984 vs. \(2 (187) = 374\).
time = 65.68, size = 1984, normalized size = 8.03 \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)^2*(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

1/192*(84*sqrt(2)*(d^10/b^4)^(1/4)*b*arctan((sqrt(d^16 + 4*sqrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a) -
2*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x + a
))*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(2*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(d^10/b^4)*b^2*d^3 + (sqrt(2)*(d^
10/b^4)^(1/4)*b*d^5*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^10/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x +
a)/cos(b*x + a))) - (sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x
+ a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(2*d^16*cos(b*x + a)^2 - d^16))*cos(b*x + a) + 84*sqrt(
2)*(d^10/b^4)^(1/4)*b*arctan(-(sqrt(d^16 + 4*sqrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*(d
^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(
b*x + a)/cos(b*x + a)))*(2*d^8*cos(b*x + a)*sin(b*x + a) + sqrt(d^10/b^4)*b^2*d^3 - (sqrt(2)*(d^10/b^4)^(1/4)*
b*d^5*cos(b*x + a)*sin(b*x + a) + sqrt(2)*(d^10/b^4)^(3/4)*b^3*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a
))) + (sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x
+ a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/(2*d^16*cos(b*x + a)^2 - d^16))*cos(b*x + a) - 84*sqrt(2)*(d^10/b^4)^
(1/4)*b*arctan(-1/2*(2*d^16*sin(b*x + a) - sqrt(d^16 + 4*sqrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a) + 2
*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x + a))
*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^5*sin(b*x + a) + sqrt(2)*(d^10/b^4)^(3/4)*b^
3*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d^11*cos(b*x + a)^3 - b^2*d^11*cos(b*x + a))*sqrt(d
^10/b^4) + (sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*sin(b*x + a) + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a))*sqrt
(d*sin(b*x + a)/cos(b*x + a)))/((2*d^16*cos(b*x + a)^2 - d^16)*sin(b*x + a)))*cos(b*x + a) - 84*sqrt(2)*(d^10/
b^4)^(1/4)*b*arctan(1/2*(2*d^16*sin(b*x + a) + sqrt(d^16 + 4*sqrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a)
- 2*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x +
a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^5*sin(b*x + a) + sqrt(2)*(d^10/b^4)^(3/4
)*b^3*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d^11*cos(b*x + a)^3 - b^2*d^11*cos(b*x + a))*sq
rt(d^10/b^4) - (sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*sin(b*x + a) + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a))*
sqrt(d*sin(b*x + a)/cos(b*x + a)))/((2*d^16*cos(b*x + a)^2 - d^16)*sin(b*x + a)))*cos(b*x + a) + 21*sqrt(2)*(d
^10/b^4)^(1/4)*b*cos(b*x + a)*log(117649*d^16 + 470596*sqrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a) + 235
298*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x +
a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 21*sqrt(2)*(d^10/b^4)^(1/4)*b*cos(b*x + a)*log(117649*d^16 + 470596*s
qrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a) - 235298*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sq
rt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) + 21*sqrt(2)*(d^1
0/b^4)^(1/4)*b*cos(b*x + a)*log(117649/16*d^16 + 117649/4*sqrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a) +
117649/8*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x + a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b
*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) - 21*sqrt(2)*(d^10/b^4)^(1/4)*b*cos(b*x + a)*log(117649/16*d^16 +
117649/4*sqrt(d^10/b^4)*b^2*d^11*cos(b*x + a)*sin(b*x + a) - 117649/8*(sqrt(2)*(d^10/b^4)^(1/4)*b*d^13*cos(b*x
+ a)^2 + sqrt(2)*(d^10/b^4)^(3/4)*b^3*d^8*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a))) + 32*
(3*d^2*cos(b*x + a)^2 + 4*d^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))*sin(b*x + a))/(b*cos(b*x + a))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.52, size = 252, normalized size = 1.02 \begin {gather*} \frac {1}{48} \, {\left (\frac {24 \, \sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b} - \frac {42 \, \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 {42 \, \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 {21 \, \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 {21 \, \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 {32 \, \sqrt {d \tan \left (b x + a\right )} \tan \left (b x + a\right )}{b}\right )} d^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(24*sqrt(d*tan(b*x + a))*d^2*tan(b*x + a)/((d^2*tan(b*x + a)^2 + d^2)*b) - 42*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) - 42*sqrt(2)*abs(d)^(3/2)*arc
tan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d)))/(b*d) + 21*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) - 21*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) + 32*sqrt(d*tan(b*x + a))*tan(b*x +
a)/b)*d^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\sin \left (a+b\,x\right )}^2\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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