3.28 \(\int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx\)

Optimal. Leaf size=209 \[ -\frac {\left (-4 c (a+2 c)+b^2+4 b c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{16 c^{3/2} e}+\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}+\frac {\sqrt {a-b+c} \tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e} \]

[Out]

-1/16*(b^2+4*b*c-4*c*(a+2*c))*arctanh(1/2*(b+2*c*tan(e*x+d)^2)/c^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)
)/c^(3/2)/e+1/2*arctanh(1/2*(2*a-b+(b-2*c)*tan(e*x+d)^2)/(a-b+c)^(1/2)/(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)
)*(a-b+c)^(1/2)/e+1/8*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*(b-4*c+2*c*tan(e*x+d)^2)/c/e

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Rubi [A]  time = 0.35, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3700, 1251, 814, 843, 621, 206, 724} \[ -\frac {\left (-4 c (a+2 c)+b^2+4 b c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{16 c^{3/2} e}+\frac {\left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}+\frac {\sqrt {a-b+c} \tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[Tan[d + e*x]^3*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]

[Out]

(Sqrt[a - b + c]*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d + e*x]^2 + c
*Tan[d + e*x]^4])])/(2*e) - ((b^2 + 4*b*c - 4*c*(a + 2*c))*ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[a
+ b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])])/(16*c^(3/2)*e) + ((b - 4*c + 2*c*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d +
e*x]^2 + c*Tan[d + e*x]^4])/(8*c*e)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

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

Rule 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 3700

Int[tan[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*((f_.)*tan[(d_.) + (e_.)*(x_)])^(n_.) + (c_.)*((f_.)*tan[(d_.
) + (e_.)*(x_)])^(n2_.))^(p_), x_Symbol] :> Dist[f/e, Subst[Int[((x/f)^m*(a + b*x^n + c*x^(2*n))^p)/(f^2 + x^2
), x], x, f*Tan[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \tan ^3(d+e x) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^3 \sqrt {a+b x^2+c x^4}}{1+x^2} \, dx,x,\tan (d+e x)\right )}{e}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x \sqrt {a+b x+c x^2}}{1+x} \, dx,x,\tan ^2(d+e x)\right )}{2 e}\\ &=\frac {\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (b^2+4 a c-4 b c\right )+\frac {1}{2} \left (b^2+4 b c-4 c (a+2 c)\right ) x}{(1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{8 c e}\\ &=\frac {\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}-\frac {(a-b+c) \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{2 e}-\frac {\left (b^2+4 b c-4 c (a+2 c)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,\tan ^2(d+e x)\right )}{16 c e}\\ &=\frac {\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}+\frac {(a-b+c) \operatorname {Subst}\left (\int \frac {1}{4 a-4 b+4 c-x^2} \, dx,x,\frac {2 a-b-(-b+2 c) \tan ^2(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{e}-\frac {\left (b^2+4 b c-4 c (a+2 c)\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c \tan ^2(d+e x)}{\sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{8 c e}\\ &=\frac {\sqrt {a-b+c} \tanh ^{-1}\left (\frac {2 a-b+(b-2 c) \tan ^2(d+e x)}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{2 e}-\frac {\left (b^2+4 b c-4 c (a+2 c)\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )}{16 c^{3/2} e}+\frac {\left (b-4 c+2 c \tan ^2(d+e x)\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{8 c e}\\ \end {align*}

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Mathematica [A]  time = 1.57, size = 208, normalized size = 1.00 \[ \frac {-\left (-4 c (a+2 c)+b^2+4 b c\right ) \tanh ^{-1}\left (\frac {b+2 c \tan ^2(d+e x)}{2 \sqrt {c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+8 c^{3/2} \sqrt {a-b+c} \tanh ^{-1}\left (\frac {2 a+(b-2 c) \tan ^2(d+e x)-b}{2 \sqrt {a-b+c} \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}\right )+2 \sqrt {c} \left (b+2 c \tan ^2(d+e x)-4 c\right ) \sqrt {a+b \tan ^2(d+e x)+c \tan ^4(d+e x)}}{16 c^{3/2} e} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[d + e*x]^3*Sqrt[a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4],x]

[Out]

(8*c^(3/2)*Sqrt[a - b + c]*ArcTanh[(2*a - b + (b - 2*c)*Tan[d + e*x]^2)/(2*Sqrt[a - b + c]*Sqrt[a + b*Tan[d +
e*x]^2 + c*Tan[d + e*x]^4])] - (b^2 + 4*b*c - 4*c*(a + 2*c))*ArcTanh[(b + 2*c*Tan[d + e*x]^2)/(2*Sqrt[c]*Sqrt[
a + b*Tan[d + e*x]^2 + c*Tan[d + e*x]^4])] + 2*Sqrt[c]*(b - 4*c + 2*c*Tan[d + e*x]^2)*Sqrt[a + b*Tan[d + e*x]^
2 + c*Tan[d + e*x]^4])/(16*c^(3/2)*e)

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fricas [A]  time = 6.37, size = 1199, normalized size = 5.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^3,x, algorithm="fricas")

[Out]

[1/32*(8*sqrt(a - b + c)*c^2*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*
c)*tan(e*x + d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(
a - b + c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2 + 1)) - (b^2 - 4*(a - b)*c - 8*c^
2)*sqrt(c)*log(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*x + d)^2 + b^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2
+ a)*(2*c*tan(e*x + d)^2 + b)*sqrt(c) + 4*a*c) + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c^2*tan(e*
x + d)^2 + b*c - 4*c^2))/(c^2*e), 1/16*(4*sqrt(a - b + c)*c^2*log(((b^2 + 4*(a - 2*b)*c + 8*c^2)*tan(e*x + d)^
4 + 2*(4*a*b - 3*b^2 - 4*(a - b)*c)*tan(e*x + d)^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c
)*tan(e*x + d)^2 + 2*a - b)*sqrt(a - b + c) + 8*a^2 - 8*a*b + b^2 + 4*a*c)/(tan(e*x + d)^4 + 2*tan(e*x + d)^2
+ 1)) + (b^2 - 4*(a - b)*c - 8*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c*tan
(e*x + d)^2 + b)*sqrt(-c)/(c^2*tan(e*x + d)^4 + b*c*tan(e*x + d)^2 + a*c)) + 2*sqrt(c*tan(e*x + d)^4 + b*tan(e
*x + d)^2 + a)*(2*c^2*tan(e*x + d)^2 + b*c - 4*c^2))/(c^2*e), 1/32*(16*sqrt(-a + b - c)*c^2*arctan(-1/2*sqrt(c
*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b)*c + c^
2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*tan(e*x + d)^2 + a^2 - a*b + a*c)) - (b^2 - 4*(a - b)*c - 8*c^2)*sqrt(c)
*log(8*c^2*tan(e*x + d)^4 + 8*b*c*tan(e*x + d)^2 + b^2 + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c*
tan(e*x + d)^2 + b)*sqrt(c) + 4*a*c) + 4*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c^2*tan(e*x + d)^2 +
b*c - 4*c^2))/(c^2*e), 1/16*(8*sqrt(-a + b - c)*c^2*arctan(-1/2*sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)
*((b - 2*c)*tan(e*x + d)^2 + 2*a - b)*sqrt(-a + b - c)/(((a - b)*c + c^2)*tan(e*x + d)^4 + (a*b - b^2 + b*c)*t
an(e*x + d)^2 + a^2 - a*b + a*c)) + (b^2 - 4*(a - b)*c - 8*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*tan(e*x + d)^4 + b*
tan(e*x + d)^2 + a)*(2*c*tan(e*x + d)^2 + b)*sqrt(-c)/(c^2*tan(e*x + d)^4 + b*c*tan(e*x + d)^2 + a*c)) + 2*sqr
t(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*(2*c^2*tan(e*x + d)^2 + b*c - 4*c^2))/(c^2*e)]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^3,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.44, size = 467, normalized size = 2.23 \[ \frac {\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\, \left (\tan ^{2}\left (e x +d \right )\right )}{4 e}+\frac {\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\, b}{8 e c}+\frac {\ln \left (\frac {c \left (\tan ^{2}\left (e x +d \right )\right )+\frac {b}{2}}{\sqrt {c}}+\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\right ) a}{4 e \sqrt {c}}-\frac {\ln \left (\frac {c \left (\tan ^{2}\left (e x +d \right )\right )+\frac {b}{2}}{\sqrt {c}}+\sqrt {a +b \left (\tan ^{2}\left (e x +d \right )\right )+c \left (\tan ^{4}\left (e x +d \right )\right )}\right ) b^{2}}{16 e \,c^{\frac {3}{2}}}-\frac {\sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}}{2 e}-\frac {\ln \left (\frac {\frac {b}{2}-c +\left (1+\tan ^{2}\left (e x +d \right )\right ) c}{\sqrt {c}}+\sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}\right ) b}{4 e \sqrt {c}}+\frac {\ln \left (\frac {\frac {b}{2}-c +\left (1+\tan ^{2}\left (e x +d \right )\right ) c}{\sqrt {c}}+\sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}\right ) \sqrt {c}}{2 e}+\frac {\sqrt {a -b +c}\, \ln \left (\frac {2 a -2 b +2 c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+2 \sqrt {a -b +c}\, \sqrt {\left (1+\tan ^{2}\left (e x +d \right )\right )^{2} c +\left (b -2 c \right ) \left (1+\tan ^{2}\left (e x +d \right )\right )+a -b +c}}{1+\tan ^{2}\left (e x +d \right )}\right )}{2 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^3,x)

[Out]

1/4/e*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^2+1/8/e/c*(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*b+1
/4/e/c^(1/2)*ln((c*tan(e*x+d)^2+1/2*b)/c^(1/2)+(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))*a-1/16/e/c^(3/2)*ln((c
*tan(e*x+d)^2+1/2*b)/c^(1/2)+(a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2))*b^2-1/2/e*((1+tan(e*x+d)^2)^2*c+(b-2*c)*
(1+tan(e*x+d)^2)+a-b+c)^(1/2)-1/4/e*ln((1/2*b-c+(1+tan(e*x+d)^2)*c)/c^(1/2)+((1+tan(e*x+d)^2)^2*c+(b-2*c)*(1+t
an(e*x+d)^2)+a-b+c)^(1/2))/c^(1/2)*b+1/2/e*ln((1/2*b-c+(1+tan(e*x+d)^2)*c)/c^(1/2)+((1+tan(e*x+d)^2)^2*c+(b-2*
c)*(1+tan(e*x+d)^2)+a-b+c)^(1/2))*c^(1/2)+1/2/e*(a-b+c)^(1/2)*ln((2*a-2*b+2*c+(b-2*c)*(1+tan(e*x+d)^2)+2*(a-b+
c)^(1/2)*((1+tan(e*x+d)^2)^2*c+(b-2*c)*(1+tan(e*x+d)^2)+a-b+c)^(1/2))/(1+tan(e*x+d)^2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c \tan \left (e x + d\right )^{4} + b \tan \left (e x + d\right )^{2} + a} \tan \left (e x + d\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(e*x+d)^2+c*tan(e*x+d)^4)^(1/2)*tan(e*x+d)^3,x, algorithm="maxima")

[Out]

integrate(sqrt(c*tan(e*x + d)^4 + b*tan(e*x + d)^2 + a)*tan(e*x + d)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (d+e\,x\right )}^3\,\sqrt {c\,{\mathrm {tan}\left (d+e\,x\right )}^4+b\,{\mathrm {tan}\left (d+e\,x\right )}^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d + e*x)^3*(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2),x)

[Out]

int(tan(d + e*x)^3*(a + b*tan(d + e*x)^2 + c*tan(d + e*x)^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan ^{2}{\left (d + e x \right )} + c \tan ^{4}{\left (d + e x \right )}} \tan ^{3}{\left (d + e x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(e*x+d)**2+c*tan(e*x+d)**4)**(1/2)*tan(e*x+d)**3,x)

[Out]

Integral(sqrt(a + b*tan(d + e*x)**2 + c*tan(d + e*x)**4)*tan(d + e*x)**3, x)

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