3.1229 \(\int (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=209 \[ \frac{2 b \left (3 a^2-b^2\right ) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac{(-b+i a)^3 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{(b+i a)^3 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

[Out]

((I*a + b)^3*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f - ((I*a - b)^3*Sqrt[c + I*d]*Arc
Tanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*b*(3*a^2 - b^2)*Sqrt[c + d*Tan[e + f*x]])/f - (4*b^2*(b*c
- 6*a*d)*(c + d*Tan[e + f*x])^(3/2))/(15*d^2*f) + (2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2))/(5*
d*f)

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Rubi [A]  time = 0.561126, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3566, 3630, 3528, 3539, 3537, 63, 208} \[ \frac{2 b \left (3 a^2-b^2\right ) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac{(-b+i a)^3 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{(b+i a)^3 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]],x]

[Out]

((I*a + b)^3*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f - ((I*a - b)^3*Sqrt[c + I*d]*Arc
Tanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*b*(3*a^2 - b^2)*Sqrt[c + d*Tan[e + f*x]])/f - (4*b^2*(b*c
- 6*a*d)*(c + d*Tan[e + f*x])^(3/2))/(15*d^2*f) + (2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2))/(5*
d*f)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rule 3537

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int (a+b \tan (e+f x))^3 \sqrt{c+d \tan (e+f x)} \, dx &=\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \int \sqrt{c+d \tan (e+f x)} \left (\frac{1}{2} \left (5 a^3 d-2 b^2 \left (b c+\frac{3 a d}{2}\right )\right )+\frac{5}{2} b \left (3 a^2-b^2\right ) d \tan (e+f x)-b^2 (b c-6 a d) \tan ^2(e+f x)\right ) \, dx}{5 d}\\ &=-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \int \sqrt{c+d \tan (e+f x)} \left (\frac{5}{2} a \left (a^2-3 b^2\right ) d+\frac{5}{2} b \left (3 a^2-b^2\right ) d \tan (e+f x)\right ) \, dx}{5 d}\\ &=\frac{2 b \left (3 a^2-b^2\right ) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{2 \int \frac{\frac{5}{2} d \left (a^3 c-3 a b^2 c-3 a^2 b d+b^3 d\right )+\frac{5}{2} d \left (3 a^2 b c-b^3 c+a^3 d-3 a b^2 d\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{5 d}\\ &=\frac{2 b \left (3 a^2-b^2\right ) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}+\frac{1}{2} \left ((a-i b)^3 (c-i d)\right ) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx+\frac{1}{2} \left ((a+i b)^3 (c+i d)\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx\\ &=\frac{2 b \left (3 a^2-b^2\right ) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac{\left (i (a+i b)^3 (c+i d)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 f}+\frac{\left ((a-i b)^3 (i c+d)\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 f}\\ &=\frac{2 b \left (3 a^2-b^2\right ) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac{\left ((a+i b)^3 (c+i d)\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}-\frac{\left ((i a+b)^3 (i c+d)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{d f}\\ &=\frac{(i a+b)^3 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f}-\frac{(i a-b)^3 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f}+\frac{2 b \left (3 a^2-b^2\right ) \sqrt{c+d \tan (e+f x)}}{f}-\frac{4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac{2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}\\ \end{align*}

Mathematica [A]  time = 2.17455, size = 194, normalized size = 0.93 \[ \frac{\frac{2 b \sqrt{c+d \tan (e+f x)} \left (45 a^2 d^2+b d (15 a d+b c) \tan (e+f x)+15 a b c d-b^2 \left (2 c^2+15 d^2\right )+3 b^2 d^2 \tan ^2(e+f x)\right )}{d^2}-15 i (a-i b)^3 \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )+15 i (a+i b)^3 \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{15 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^3*Sqrt[c + d*Tan[e + f*x]],x]

[Out]

((-15*I)*(a - I*b)^3*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + (15*I)*(a + I*b)^3*Sqrt[c
+ I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + (2*b*Sqrt[c + d*Tan[e + f*x]]*(15*a*b*c*d + 45*a^2*d
^2 - b^2*(2*c^2 + 15*d^2) + b*d*(b*c + 15*a*d)*Tan[e + f*x] + 3*b^2*d^2*Tan[e + f*x]^2))/d^2)/(15*f)

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Maple [B]  time = 0.077, size = 2073, normalized size = 9.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^
(1/2)+2*c)^(1/2)*a^3*c+3/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*
x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*b^2-3/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e
))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a*b^2+1/4/f/d*ln((c+d*tan(f*x+e))^(1/2)
*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a
^3-1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2
)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a^3+1/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c
)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^3*c+3/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*
tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a^2*b*c+3/f/(2*(c^2+d^2)^(1/2)
-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^
2+d^2)^(1/2)*a^2*b-3/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^
(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*a^2*b-3/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+
d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a^2*b*c+2/5/f/d^2*b^3*(c+d*tan(
f*x+e))^(5/2)+1/4/f*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2
*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^3-1/4/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(
c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*b^3+6/f*(c+d*tan(f*x+e))^(1/2)*a^2*b-2/f*b^3*(c+d*tan(f*x+e))^(1
/2)-1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^
2+d^2)^(1/2)-2*c)^(1/2))*a^3+1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*ta
n(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b^3*c+3/4/f*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^
(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2*b-1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*ar
ctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b
^3-2/3/f/d^2*(c+d*tan(f*x+e))^(3/2)*b^3*c+2/f/d*(c+d*tan(f*x+e))^(3/2)*a*b^2+1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/
2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a^3-3/4/f*ln
(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^
(1/2)*a^2*b+1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/
(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)*b^3-1/f/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e)
)^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*b^3*c-3/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*
(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*
b^2+3/4/f/d*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^
2)^(1/2)+2*c)^(1/2)*a*b^2*c+3/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^
2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*a*b^2-3/4/f/d*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(
1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a*b^2*c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{3} \sqrt{d \tan \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (e + f x \right )}\right )^{3} \sqrt{c + d \tan{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{3} \sqrt{d \tan \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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