3.514 \(\int (a+b \tan (d+e x)) (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x))^{3/2} \, dx\)

Optimal. Leaf size=284 \[ \frac {b \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{3 e}+\frac {\left (a^2+b^2\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{2 e (a \tan (d+e x)+b)}-\frac {2 a^4 b x \left (a^2+b^2\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{\left (a^2 \tan (d+e x)+a b\right )^3}+\frac {a^4 b \left (a^2+b^2\right ) \tan (d+e x) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{e \left (a^2 \tan (d+e x)+a b\right )^3}+\frac {\left (a^4-b^4\right ) \log (\cos (d+e x)) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{e (a \tan (d+e x)+b)^3} \]

[Out]

1/3*b*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(3/2)/e+(a^4-b^4)*ln(cos(e*x+d))*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e
*x+d)^2)^(3/2)/e/(b+a*tan(e*x+d))^3+1/2*(a^2+b^2)*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(3/2)/e/(b+a*tan(e*x
+d))-2*a^4*b*(a^2+b^2)*x*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(3/2)/(a*b+a^2*tan(e*x+d))^3+a^4*b*(a^2+b^2)*
tan(e*x+d)*(b^2+2*a*b*tan(e*x+d)+a^2*tan(e*x+d)^2)^(3/2)/e/(a*b+a^2*tan(e*x+d))^3

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Rubi [A]  time = 0.23, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3710, 3528, 12, 3525, 3475} \[ -\frac {2 a^4 b x \left (a^2+b^2\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{\left (a^2 \tan (d+e x)+a b\right )^3}+\frac {a^4 b \left (a^2+b^2\right ) \tan (d+e x) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{e \left (a^2 \tan (d+e x)+a b\right )^3}+\frac {b \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{3 e}+\frac {\left (a^2+b^2\right ) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{2 e (a \tan (d+e x)+b)}+\frac {\left (a^4-b^4\right ) \log (\cos (d+e x)) \left (a^2 \tan ^2(d+e x)+2 a b \tan (d+e x)+b^2\right )^{3/2}}{e (a \tan (d+e x)+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(b^2 + 2*a*b*Tan[d + e*x] + a^2*Tan[d + e*x]^2)^(3/2))/(3*e) + ((a^4 - b^4)*Log[Cos[d + e*x]]*(b^2 + 2*a*b*
Tan[d + e*x] + a^2*Tan[d + e*x]^2)^(3/2))/(e*(b + a*Tan[d + e*x])^3) + ((a^2 + b^2)*(b^2 + 2*a*b*Tan[d + e*x]
+ a^2*Tan[d + e*x]^2)^(3/2))/(2*e*(b + a*Tan[d + e*x])) - (2*a^4*b*(a^2 + b^2)*x*(b^2 + 2*a*b*Tan[d + e*x] + a
^2*Tan[d + e*x]^2)^(3/2))/(a*b + a^2*Tan[d + e*x])^3 + (a^4*b*(a^2 + b^2)*Tan[d + e*x]*(b^2 + 2*a*b*Tan[d + e*
x] + a^2*Tan[d + e*x]^2)^(3/2))/(e*(a*b + a^2*Tan[d + e*x])^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3525

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

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 3710

Int[((A_) + (B_.)*tan[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*tan[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[(a + b*Tan[d + e*x] + c*Tan[d + e*x]^2)^n/(b + 2*c*Tan[d + e*x])^(2*n), Int[(A +
B*Tan[d + e*x])*(b + 2*c*Tan[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0
] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (a+b \tan (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2} \, dx &=\frac {\left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2} \int \left (2 a b+2 a^2 \tan (d+e x)\right )^3 (a+b \tan (d+e x)) \, dx}{\left (2 a b+2 a^2 \tan (d+e x)\right )^3}\\ &=\frac {b \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{3 e}+\frac {\left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2} \int 2 a \left (a^2+b^2\right ) \tan (d+e x) \left (2 a b+2 a^2 \tan (d+e x)\right )^2 \, dx}{\left (2 a b+2 a^2 \tan (d+e x)\right )^3}\\ &=\frac {b \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{3 e}+\frac {\left (2 a \left (a^2+b^2\right ) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}\right ) \int \tan (d+e x) \left (2 a b+2 a^2 \tan (d+e x)\right )^2 \, dx}{\left (2 a b+2 a^2 \tan (d+e x)\right )^3}\\ &=\frac {b \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{3 e}+\frac {\left (a^2+b^2\right ) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{2 e (b+a \tan (d+e x))}+\frac {\left (2 a \left (a^2+b^2\right ) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}\right ) \int \left (2 a b+2 a^2 \tan (d+e x)\right ) \left (-2 a^2+2 a b \tan (d+e x)\right ) \, dx}{\left (2 a b+2 a^2 \tan (d+e x)\right )^3}\\ &=\frac {b \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{3 e}+\frac {\left (a^2+b^2\right ) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{2 e (b+a \tan (d+e x))}-\frac {2 a^4 b \left (a^2+b^2\right ) x \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \tan (d+e x)\right )^3}+\frac {a^4 b \left (a^2+b^2\right ) \tan (d+e x) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{e \left (a b+a^2 \tan (d+e x)\right )^3}-\frac {\left (8 a^3 \left (a^2-b^2\right ) \left (a^2+b^2\right ) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}\right ) \int \tan (d+e x) \, dx}{\left (2 a b+2 a^2 \tan (d+e x)\right )^3}\\ &=\frac {b \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{3 e}+\frac {\left (a^4-b^4\right ) \log (\cos (d+e x)) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{e (b+a \tan (d+e x))^3}+\frac {\left (a^2+b^2\right ) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{2 e (b+a \tan (d+e x))}-\frac {2 a^4 b \left (a^2+b^2\right ) x \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{\left (a b+a^2 \tan (d+e x)\right )^3}+\frac {a^4 b \left (a^2+b^2\right ) \tan (d+e x) \left (b^2+2 a b \tan (d+e x)+a^2 \tan ^2(d+e x)\right )^{3/2}}{e \left (a b+a^2 \tan (d+e x)\right )^3}\\ \end {align*}

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Mathematica [C]  time = 1.33, size = 147, normalized size = 0.52 \[ \frac {\sqrt {(a \tan (d+e x)+b)^2} \left (2 a^3 b \tan ^3(d+e x)+3 a^2 \left (a^2+3 b^2\right ) \tan ^2(d+e x)+6 a b \left (2 a^2+3 b^2\right ) \tan (d+e x)-3 \left (a^2+b^2\right ) \left ((a-i b)^2 \log (-\tan (d+e x)+i)+(a+i b)^2 \log (\tan (d+e x)+i)\right )\right )}{6 e (a \tan (d+e x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(b + a*Tan[d + e*x])^2]*(-3*(a^2 + b^2)*((a - I*b)^2*Log[I - Tan[d + e*x]] + (a + I*b)^2*Log[I + Tan[d +
e*x]]) + 6*a*b*(2*a^2 + 3*b^2)*Tan[d + e*x] + 3*a^2*(a^2 + 3*b^2)*Tan[d + e*x]^2 + 2*a^3*b*Tan[d + e*x]^3))/(
6*e*(b + a*Tan[d + e*x]))

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fricas [A]  time = 2.83, size = 102, normalized size = 0.36 \[ \frac {2 \, a^{3} b \tan \left (e x + d\right )^{3} - 12 \, {\left (a^{3} b + a b^{3}\right )} e x + 3 \, {\left (a^{4} + 3 \, a^{2} b^{2}\right )} \tan \left (e x + d\right )^{2} + 3 \, {\left (a^{4} - b^{4}\right )} \log \left (\frac {1}{\tan \left (e x + d\right )^{2} + 1}\right ) + 6 \, {\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \tan \left (e x + d\right )}{6 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a^3*b*tan(e*x + d)^3 - 12*(a^3*b + a*b^3)*e*x + 3*(a^4 + 3*a^2*b^2)*tan(e*x + d)^2 + 3*(a^4 - b^4)*log(
1/(tan(e*x + d)^2 + 1)) + 6*(2*a^3*b + 3*a*b^3)*tan(e*x + d))/e

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giac [B]  time = 2.44, size = 1751, normalized size = 6.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*a^3*b*x*e*sgn(a*tan(x*e + d) + b)*tan(x*e)^3*tan(d)^3 + 12*a*b^3*x*e*sgn(a*tan(x*e + d) + b)*tan(x*e)
^3*tan(d)^3 - 3*a^4*log(4*(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*ta
n(x*e)*tan(d) + 1)/(tan(d)^2 + 1))*sgn(a*tan(x*e + d) + b)*tan(x*e)^3*tan(d)^3 + 3*b^4*log(4*(tan(x*e)^4*tan(d
)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1)/(tan(d)^2 + 1))*sgn(a*ta
n(x*e + d) + b)*tan(x*e)^3*tan(d)^3 - 36*a^3*b*x*e*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d)^2 - 36*a*b^3*x*e*
sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d)^2 - 3*a^4*sgn(a*tan(x*e + d) + b)*tan(x*e)^3*tan(d)^3 - 9*a^2*b^2*sg
n(a*tan(x*e + d) + b)*tan(x*e)^3*tan(d)^3 + 9*a^4*log(4*(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^
2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1)/(tan(d)^2 + 1))*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d)^2 -
9*b^4*log(4*(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d)
+ 1)/(tan(d)^2 + 1))*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d)^2 + 12*a^3*b*sgn(a*tan(x*e + d) + b)*tan(x*e)^
3*tan(d)^2 + 18*a*b^3*sgn(a*tan(x*e + d) + b)*tan(x*e)^3*tan(d)^2 + 12*a^3*b*sgn(a*tan(x*e + d) + b)*tan(x*e)^
2*tan(d)^3 + 18*a*b^3*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d)^3 + 36*a^3*b*x*e*sgn(a*tan(x*e + d) + b)*tan(x
*e)*tan(d) + 36*a*b^3*x*e*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(d) - 3*a^4*sgn(a*tan(x*e + d) + b)*tan(x*e)^3*t
an(d) - 9*a^2*b^2*sgn(a*tan(x*e + d) + b)*tan(x*e)^3*tan(d) + 3*a^4*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d)^
2 + 9*a^2*b^2*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d)^2 - 3*a^4*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(d)^3 -
9*a^2*b^2*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(d)^3 + 2*a^3*b*sgn(a*tan(x*e + d) + b)*tan(x*e)^3 - 9*a^4*log(4
*(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1)/(tan(d
)^2 + 1))*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(d) + 9*b^4*log(4*(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + t
an(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1)/(tan(d)^2 + 1))*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(
d) - 18*a^3*b*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d) - 36*a*b^3*sgn(a*tan(x*e + d) + b)*tan(x*e)^2*tan(d) -
18*a^3*b*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(d)^2 - 36*a*b^3*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(d)^2 + 2*a
^3*b*sgn(a*tan(x*e + d) + b)*tan(d)^3 - 12*a^3*b*x*e*sgn(a*tan(x*e + d) + b) - 12*a*b^3*x*e*sgn(a*tan(x*e + d)
+ b) + 3*a^4*sgn(a*tan(x*e + d) + b)*tan(x*e)^2 + 9*a^2*b^2*sgn(a*tan(x*e + d) + b)*tan(x*e)^2 - 3*a^4*sgn(a*
tan(x*e + d) + b)*tan(x*e)*tan(d) - 9*a^2*b^2*sgn(a*tan(x*e + d) + b)*tan(x*e)*tan(d) + 3*a^4*sgn(a*tan(x*e +
d) + b)*tan(d)^2 + 9*a^2*b^2*sgn(a*tan(x*e + d) + b)*tan(d)^2 + 3*a^4*log(4*(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^
3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d) + 1)/(tan(d)^2 + 1))*sgn(a*tan(x*e + d) + b) -
3*b^4*log(4*(tan(x*e)^4*tan(d)^2 - 2*tan(x*e)^3*tan(d) + tan(x*e)^2*tan(d)^2 + tan(x*e)^2 - 2*tan(x*e)*tan(d)
+ 1)/(tan(d)^2 + 1))*sgn(a*tan(x*e + d) + b) + 12*a^3*b*sgn(a*tan(x*e + d) + b)*tan(x*e) + 18*a*b^3*sgn(a*tan
(x*e + d) + b)*tan(x*e) + 12*a^3*b*sgn(a*tan(x*e + d) + b)*tan(d) + 18*a*b^3*sgn(a*tan(x*e + d) + b)*tan(d) +
3*a^4*sgn(a*tan(x*e + d) + b) + 9*a^2*b^2*sgn(a*tan(x*e + d) + b))/(e*tan(x*e)^3*tan(d)^3 - 3*e*tan(x*e)^2*tan
(d)^2 + 3*e*tan(x*e)*tan(d) - e)

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maple [A]  time = 0.31, size = 158, normalized size = 0.56 \[ -\frac {\left (b^{2}+2 a b \tan \left (e x +d \right )+a^{2} \left (\tan ^{2}\left (e x +d \right )\right )\right )^{\frac {3}{2}} \left (-2 \left (\tan ^{3}\left (e x +d \right )\right ) a^{3} b -3 \left (\tan ^{2}\left (e x +d \right )\right ) a^{4}-9 \left (\tan ^{2}\left (e x +d \right )\right ) a^{2} b^{2}+3 \ln \left (1+\tan ^{2}\left (e x +d \right )\right ) a^{4}-3 \ln \left (1+\tan ^{2}\left (e x +d \right )\right ) b^{4}+12 \arctan \left (\tan \left (e x +d \right )\right ) a^{3} b +12 \arctan \left (\tan \left (e x +d \right )\right ) a \,b^{3}-12 \tan \left (e x +d \right ) a^{3} b -18 \tan \left (e x +d \right ) a \,b^{3}\right )}{6 e \left (b +a \tan \left (e x +d \right )\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/e*((b+a*tan(e*x+d))^2)^(3/2)*(-2*tan(e*x+d)^3*a^3*b-3*tan(e*x+d)^2*a^4-9*tan(e*x+d)^2*a^2*b^2+3*ln(1+tan(
e*x+d)^2)*a^4-3*ln(1+tan(e*x+d)^2)*b^4+12*arctan(tan(e*x+d))*a^3*b+12*arctan(tan(e*x+d))*a*b^3-12*tan(e*x+d)*a
^3*b-18*tan(e*x+d)*a*b^3)/(b+a*tan(e*x+d))^3

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maxima [A]  time = 0.43, size = 166, normalized size = 0.58 \[ \frac {3 \, {\left (a^{3} \tan \left (e x + d\right )^{2} + 6 \, a^{2} b \tan \left (e x + d\right ) - 2 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (e x + d\right )} - {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right )\right )} a + {\left (2 \, a^{3} \tan \left (e x + d\right )^{3} + 9 \, a^{2} b \tan \left (e x + d\right )^{2} + 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} {\left (e x + d\right )} - 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \log \left (\tan \left (e x + d\right )^{2} + 1\right ) - 6 \, {\left (a^{3} - 3 \, a b^{2}\right )} \tan \left (e x + d\right )\right )} b}{6 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(a^3*tan(e*x + d)^2 + 6*a^2*b*tan(e*x + d) - 2*(3*a^2*b - b^3)*(e*x + d) - (a^3 - 3*a*b^2)*log(tan(e*x
+ d)^2 + 1))*a + (2*a^3*tan(e*x + d)^3 + 9*a^2*b*tan(e*x + d)^2 + 6*(a^3 - 3*a*b^2)*(e*x + d) - 3*(3*a^2*b - b
^3)*log(tan(e*x + d)^2 + 1) - 6*(a^3 - 3*a*b^2)*tan(e*x + d))*b)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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