3.6.6 \(\int (a+b \sin ^2(e+f x))^{3/2} \tan ^2(e+f x) \, dx\) [506]

Optimal. Leaf size=222 \[ \frac {4 b \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {(7 a+8 b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {4 a (a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{f} \]

[Out]

4/3*b*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/f-1/3*(7*a+8*b)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*sec(f*
x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e)^2)^(1/2)/f/(1+b*sin(f*x+e)^2/a)^(1/2)+4/3*a*(a+b)*EllipticF(sin(f*x+
e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/f/(a+b*sin(f*x+e)^2)^(1/2)+(a+b*si
n(f*x+e)^2)^(3/2)*tan(f*x+e)/f

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Rubi [A]
time = 0.15, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3275, 478, 542, 538, 437, 435, 432, 430} \begin {gather*} \frac {4 a (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(7 a+8 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\tan (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{f}+\frac {4 b \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*f) - ((7*a + 8*b)*Sqrt[Cos[e + f*x]^2]*EllipticE
[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) +
(4*a*(a + b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e + f*
x]^2)/a])/(3*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x])/f

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]
, Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 478

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[e^(n - 1
)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*n*(p + 1))), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

Rule 542

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(n*(p + q + 1) + 1))), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]

Rule 3275

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff^(m + 1)*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), Subst[Int[x^m*((a + b*ff^2*
x^2)^p/(1 - ff^2*x^2)^((m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^2(e+f x) \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^{3/2}}{\left (1-x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2} \left (a+4 b x^2\right )}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {4 b \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{f}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-a (3 a+4 b)-b (7 a+8 b) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 f}\\ &=\frac {4 b \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{f}+\frac {\left ((-7 a-8 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 f}+\frac {\left (4 a (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 f}\\ &=\frac {4 b \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{f}+\frac {\left ((-7 a-8 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (4 a (a+b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{3 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=\frac {4 b \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {(7 a+8 b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {4 a (a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A]
time = 1.97, size = 174, normalized size = 0.78 \begin {gather*} \frac {-8 a (7 a+8 b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+32 a (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} \left (24 a^2+40 a b+13 b^2-4 b (2 a+3 b) \cos (2 (e+f x))-b^2 \cos (4 (e+f x))\right ) \tan (e+f x)}{24 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*(7*a + 8*b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 32*a*(a + b)*Sqrt[(2*a +
b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + Sqrt[2]*(24*a^2 + 40*a*b + 13*b^2 - 4*b*(2*a + 3*b)*C
os[2*(e + f*x)] - b^2*Cos[4*(e + f*x)])*Tan[e + f*x])/(24*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(514\) vs. \(2(202)=404\).
time = 18.77, size = 515, normalized size = 2.32

method	result	size
default	\(\frac {-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{2} \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+3 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )+4 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+4 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}-7 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, a^{2}-8 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, a b}{3 \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(515\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^2*sin(f*x+e)*cos(f*x+e)^4-2*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+
e)^2)^(1/2)*b*(a+b)*cos(f*x+e)^2*sin(f*x+e)+3*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a^2+2*a*b+b^2)*sin(f
*x+e)+4*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2*(-b*co
s(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)+4*a*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(
f*x+e),(-1/a*b)^(1/2))*b*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)-7*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+
(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a^2-8*(cos(f*x+
e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*(-b*cos(f*x+e)^4+(a+b)*cos(
f*x+e)^2)^(1/2)*a*b)/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(
1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^2, x)

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Fricas [F]
time = 0.13, size = 45, normalized size = 0.20 \begin {gather*} {\rm integral}\left (-{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )^{2}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*cos(f*x + e)^2 - a - b)*sqrt(-b*cos(f*x + e)^2 + a + b)*tan(f*x + e)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)**2)**(3/2)*tan(f*x+e)**2,x)

[Out]

Integral((a + b*sin(e + f*x)**2)**(3/2)*tan(e + f*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^2*(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(tan(e + f*x)^2*(a + b*sin(e + f*x)^2)^(3/2), x)

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