3.387 \(\int \sqrt{a+b \tan ^4(c+d x)} \, dx\)

Optimal. Leaf size=650 \[ -\frac{\sqrt [4]{b} (a+b) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}}+\frac{\sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{2 \sqrt [4]{a} d \sqrt{a+b \tan ^4(c+d x)}}+\frac{\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a+b \tan ^4(c+d x)}}\right )}{2 d}+\frac{\sqrt{b} \tan (c+d x) \sqrt{a+b \tan ^4(c+d x)}}{d \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}-\frac{\sqrt [4]{a} \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{d \sqrt{a+b \tan ^4(c+d x)}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) (a+b) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}} \]

[Out]

(Sqrt[a + b]*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^4]])/(2*d) + (Sqrt[b]*Tan[c + d*x]*Sqrt
[a + b*Tan[c + d*x]^4])/(d*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)) - (a^(1/4)*b^(1/4)*EllipticE[2*ArcTan[(b^(1/4)*
Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]
*Tan[c + d*x]^2)^2])/(d*Sqrt[a + b*Tan[c + d*x]^4]) + ((Sqrt[a] - Sqrt[b])*b^(1/4)*EllipticF[2*ArcTan[(b^(1/4)
*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b
]*Tan[c + d*x]^2)^2])/(2*a^(1/4)*d*Sqrt[a + b*Tan[c + d*x]^4]) - (b^(1/4)*(a + b)*EllipticF[2*ArcTan[(b^(1/4)*
Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]
*Tan[c + d*x]^2)^2])/(2*a^(1/4)*(Sqrt[a] - Sqrt[b])*d*Sqrt[a + b*Tan[c + d*x]^4]) + ((Sqrt[a] + Sqrt[b])*(a +
b)*EllipticPi[-(Sqrt[a] - Sqrt[b])^2/(4*Sqrt[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt
[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)^2])/(4*a^(1/4)*(S
qrt[a] - Sqrt[b])*b^(1/4)*d*Sqrt[a + b*Tan[c + d*x]^4])

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Rubi [A]  time = 0.559916, antiderivative size = 650, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3661, 1209, 1198, 220, 1196, 1217, 1707} \[ \frac{\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a+b \tan ^4(c+d x)}}\right )}{2 d}+\frac{\sqrt{b} \tan (c+d x) \sqrt{a+b \tan ^4(c+d x)}}{d \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}-\frac{\sqrt [4]{b} (a+b) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}}+\frac{\sqrt [4]{b} \left (\sqrt{a}-\sqrt{b}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{2 \sqrt [4]{a} d \sqrt{a+b \tan ^4(c+d x)}}-\frac{\sqrt [4]{a} \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{d \sqrt{a+b \tan ^4(c+d x)}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) (a+b) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}} \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{4 \sqrt [4]{a} \sqrt [4]{b} d \left (\sqrt{a}-\sqrt{b}\right ) \sqrt{a+b \tan ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Tan[c + d*x]^4],x]

[Out]

(Sqrt[a + b]*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a + b*Tan[c + d*x]^4]])/(2*d) + (Sqrt[b]*Tan[c + d*x]*Sqrt
[a + b*Tan[c + d*x]^4])/(d*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)) - (a^(1/4)*b^(1/4)*EllipticE[2*ArcTan[(b^(1/4)*
Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]
*Tan[c + d*x]^2)^2])/(d*Sqrt[a + b*Tan[c + d*x]^4]) + ((Sqrt[a] - Sqrt[b])*b^(1/4)*EllipticF[2*ArcTan[(b^(1/4)
*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b
]*Tan[c + d*x]^2)^2])/(2*a^(1/4)*d*Sqrt[a + b*Tan[c + d*x]^4]) - (b^(1/4)*(a + b)*EllipticF[2*ArcTan[(b^(1/4)*
Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]
*Tan[c + d*x]^2)^2])/(2*a^(1/4)*(Sqrt[a] - Sqrt[b])*d*Sqrt[a + b*Tan[c + d*x]^4]) + ((Sqrt[a] + Sqrt[b])*(a +
b)*EllipticPi[-(Sqrt[a] - Sqrt[b])^2/(4*Sqrt[a]*Sqrt[b]), 2*ArcTan[(b^(1/4)*Tan[c + d*x])/a^(1/4)], 1/2]*(Sqrt
[a] + Sqrt[b]*Tan[c + d*x]^2)*Sqrt[(a + b*Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[b]*Tan[c + d*x]^2)^2])/(4*a^(1/4)*(S
qrt[a] - Sqrt[b])*b^(1/4)*d*Sqrt[a + b*Tan[c + d*x]^4])

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 1209

Int[((a_) + (c_.)*(x_)^4)^(p_)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[(e^2)^(-1), Int[(c*d - c*e*x^2)*(a +
c*x^4)^(p - 1), x], x] + Dist[(c*d^2 + a*e^2)/e^2, Int[(a + c*x^4)^(p - 1)/(d + e*x^2), x], x] /; FreeQ[{a, c,
d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p + 1/2, 0]

Rule 1198

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(e + d*q)/q, Int
[1/Sqrt[a + c*x^4], x], x] - Dist[e/q, Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a
, c, d, e}, x] && PosQ[c/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1217

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(c*d + a*e*q
)/(c*d^2 - a*e^2), Int[1/Sqrt[a + c*x^4], x], x] - Dist[(a*e*(e + d*q))/(c*d^2 - a*e^2), Int[(1 + q*x^2)/((d +
e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0]
&& PosQ[c/a]

Rule 1707

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]
}, -Simp[((B*d - A*e)*ArcTan[(Rt[(c*d)/e + (a*e)/d, 2]*x)/Sqrt[a + c*x^4]])/(2*d*e*Rt[(c*d)/e + (a*e)/d, 2]),
x] + Simp[((B*d + A*e)*(A + B*x^2)*Sqrt[(A^2*(a + c*x^4))/(a*(A + B*x^2)^2)]*EllipticPi[Cancel[-((B*d - A*e)^2
/(4*d*e*A*B))], 2*ArcTan[q*x], 1/2])/(4*d*e*A*q*Sqrt[a + c*x^4]), x]] /; FreeQ[{a, c, d, e, A, B}, x] && NeQ[c
*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0]

Rubi steps

\begin{align*} \int \sqrt{a+b \tan ^4(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^4}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{b-b x^2}{\sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{d}+\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\left (\sqrt{a} \sqrt{b}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (\left (\sqrt{a}-\sqrt{b}\right ) \sqrt{b}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (\sqrt{a} (a+b)\right ) \operatorname{Subst}\left (\int \frac{1+\frac{\sqrt{b} x^2}{\sqrt{a}}}{\left (1+x^2\right ) \sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{\left (\sqrt{a}-\sqrt{b}\right ) d}-\frac{\left (\sqrt{b} (a+b)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\tan (c+d x)\right )}{\left (\sqrt{a}-\sqrt{b}\right ) d}\\ &=\frac{\sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a+b} \tan (c+d x)}{\sqrt{a+b \tan ^4(c+d x)}}\right )}{2 d}+\frac{\sqrt{b} \tan (c+d x) \sqrt{a+b \tan ^4(c+d x)}}{d \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )}-\frac{\sqrt [4]{a} \sqrt [4]{b} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}}}{d \sqrt{a+b \tan ^4(c+d x)}}+\frac{\left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}}}{2 \sqrt [4]{a} d \sqrt{a+b \tan ^4(c+d x)}}-\frac{\sqrt [4]{b} (a+b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}}}{2 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{b}\right ) d \sqrt{a+b \tan ^4(c+d x)}}+\frac{\left (\sqrt{a}+\sqrt{b}\right ) (a+b) \Pi \left (-\frac{\left (\sqrt{a}-\sqrt{b}\right )^2}{4 \sqrt{a} \sqrt{b}};2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac{1}{2}\right ) \left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right ) \sqrt{\frac{a+b \tan ^4(c+d x)}{\left (\sqrt{a}+\sqrt{b} \tan ^2(c+d x)\right )^2}}}{4 \sqrt [4]{a} \left (\sqrt{a}-\sqrt{b}\right ) \sqrt [4]{b} d \sqrt{a+b \tan ^4(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.725444, size = 219, normalized size = 0.34 \[ \frac{\sqrt{\frac{b \tan ^4(c+d x)}{a}+1} \left (\left (\sqrt{a}-i \sqrt{b}\right ) \left (\left (\sqrt{b}-i \sqrt{a}\right ) \Pi \left (-\frac{i \sqrt{a}}{\sqrt{b}};\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right )\right |-1\right )-\sqrt{b} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right ),-1\right )\right )+\sqrt{a} \sqrt{b} E\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \tan (c+d x)\right )\right |-1\right )\right )}{d \sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} \sqrt{a+b \tan ^4(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*Sqrt[b]*EllipticE[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], -1] + (Sqrt[a] - I*Sqrt[b])*(-(
Sqrt[b]*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], -1]) + ((-I)*Sqrt[a] + Sqrt[b])*EllipticP
i[((-I)*Sqrt[a])/Sqrt[b], I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*Tan[c + d*x]], -1]))*Sqrt[1 + (b*Tan[c + d*x]^4)
/a])/(Sqrt[(I*Sqrt[b])/Sqrt[a]]*d*Sqrt[a + b*Tan[c + d*x]^4])

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Maple [C]  time = 0.12, size = 531, normalized size = 0.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-b/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^
(1/2)/(a+b*tan(d*x+c)^4)^(1/2)*EllipticF(tan(d*x+c)*(I/a^(1/2)*b^(1/2))^(1/2),I)+I*b^(1/2)*a^(1/2)/(I/a^(1/2)*
b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)/(a+b*tan(d*x+
c)^4)^(1/2)*EllipticF(tan(d*x+c)*(I/a^(1/2)*b^(1/2))^(1/2),I)-I*b^(1/2)*a^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I
/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)/(a+b*tan(d*x+c)^4)^(1/2)*Ellipti
cE(tan(d*x+c)*(I/a^(1/2)*b^(1/2))^(1/2),I)+a/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2
)*(1+I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)/(a+b*tan(d*x+c)^4)^(1/2)*EllipticPi(tan(d*x+c)*(I/a^(1/2)*b^(1/2))^
(1/2),I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2))+b/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I
/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*tan(d*x+c)^2)^(1/2)/(a+b*tan(d*x+c)^4)^(1/2)*Ellipti
cPi(tan(d*x+c)*(I/a^(1/2)*b^(1/2))^(1/2),I*a^(1/2)/b^(1/2),(-I/a^(1/2)*b^(1/2))^(1/2)/(I/a^(1/2)*b^(1/2))^(1/2
)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (d x + c\right )^{4} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(d*x + c)^4 + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \tan \left (d x + c\right )^{4} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*tan(d*x + c)^4 + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \tan ^{4}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (d x + c\right )^{4} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*tan(d*x + c)^4 + a), x)