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

Optimal. Leaf size=111 \[ \frac {2 b \sqrt {a+b \tan (c+d x)}}{d}-\frac {i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d} \]

[Out]

-I*(a-I*b)^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+I*(a+I*b)^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2
)/(a+I*b)^(1/2))/d+2*b*(a+b*tan(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.17, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3482, 3539, 3537, 63, 208} \[ \frac {2 b \sqrt {a+b \tan (c+d x)}}{d}-\frac {i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(3/2),x]

[Out]

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

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]

Rule 3482

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 106, normalized size = 0.95 \[ \frac {2 b \sqrt {a+b \tan (c+d x)}-i (a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+i (a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(a - I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + I*(a + I*b)^(3/2)*ArcTanh[Sqrt[a + b*T
an[c + d*x]]/Sqrt[a + I*b]] + 2*b*Sqrt[a + b*Tan[c + d*x]])/d

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fricas [B]  time = 1.21, size = 4150, normalized size = 37.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*sqrt(2)*d^5*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2
*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(3/4)*sqrt((9*a^4*b
^2 - 6*a^2*b^4 + b^6)/d^4)*arctan(((3*a^10 + 11*a^8*b^2 + 14*a^6*b^4 + 6*a^4*b^6 - a^2*b^8 - b^10)*d^4*sqrt((a
^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + (3*a^13 + 14*a^11*b^2 + 25*a^
9*b^4 + 20*a^7*b^6 + 5*a^5*b^8 - 2*a^3*b^10 - a*b^12)*d^2*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + sqrt(2)*((
3*a^4*b + 2*a^2*b^3 - b^5)*d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6
)/d^4) + 2*(3*a^7*b + 5*a^5*b^3 + a^3*b^5 - a*b^7)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))*sqrt((a^6 + 3*
a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*
a^2*b^4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)
^(3/4) + sqrt(2)*(d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) +
2*(a^3 + a*b^2)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 -
3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt(((9*a^8*b^2 +
12*a^6*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + sq
rt(2)*(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + (9*a
^8*b^3 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 +
(a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a*cos(
d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4) + (9*a^11*b^2 + 21*a^
9*b^4 + 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^10 + a*b^12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^8*b^5 + 10*a^6*b^7 - 6
*a^4*b^9 - 3*a^2*b^11 + b^13)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d
^4)^(3/4))/(9*a^14*b^2 + 39*a^12*b^4 + 61*a^10*b^6 + 35*a^8*b^8 - 5*a^6*b^10 - 11*a^4*b^12 - a^2*b^14 + b^16))
 + 4*sqrt(2)*d^5*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b
^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(3/4)*sqrt((9*a^4*b^2
 - 6*a^2*b^4 + b^6)/d^4)*arctan(-((3*a^10 + 11*a^8*b^2 + 14*a^6*b^4 + 6*a^4*b^6 - a^2*b^8 - b^10)*d^4*sqrt((a^
6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + (3*a^13 + 14*a^11*b^2 + 25*a^9
*b^4 + 20*a^7*b^6 + 5*a^5*b^8 - 2*a^3*b^10 - a*b^12)*d^2*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) - sqrt(2)*((3
*a^4*b + 2*a^2*b^3 - b^5)*d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)
/d^4) + 2*(3*a^7*b + 5*a^5*b^3 + a^3*b^5 - a*b^7)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))*sqrt((a^6 + 3*a
^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a
^2*b^4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^
(3/4) - sqrt(2)*(d^7*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4) + 2
*(a^3 + a*b^2)*d^5*sqrt((9*a^4*b^2 - 6*a^2*b^4 + b^6)/d^4))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3
*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt(((9*a^8*b^2 + 1
2*a^6*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) - sqr
t(2)*(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + (9*a^
8*b^3 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (
a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a*cos(d
*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4) + (9*a^11*b^2 + 21*a^9
*b^4 + 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^10 + a*b^12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^8*b^5 + 10*a^6*b^7 - 6*
a^4*b^9 - 3*a^2*b^11 + b^13)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^
4)^(3/4))/(9*a^14*b^2 + 39*a^12*b^4 + 61*a^10*b^6 + 35*a^8*b^8 - 5*a^6*b^10 - 11*a^4*b^12 - a^2*b^14 + b^16))
+ sqrt(2)*((a^3 - 3*a*b^2)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4) - (a^6 + 3*a^4*b^2 + 3*a^2*b^4 +
b^6)*d)*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)
/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4)*log(((9*a^8*b^2 + 12*a^6
*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + sqrt(2)*
(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + (9*a^8*b^3
 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 -
 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a*cos(d*x +
c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4) + (9*a^11*b^2 + 21*a^9*b^4
+ 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^10 + a*b^12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^8*b^5 + 10*a^6*b^7 - 6*a^4*b
^9 - 3*a^2*b^11 + b^13)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c))) - sqrt(2)*((a^3 - 3*a*b^2)*d^3*sqrt((a^6 + 3
*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4) - (a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d)*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 +
b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*((a^6
+ 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4)*log(((9*a^8*b^2 + 12*a^6*b^4 - 2*a^4*b^6 - 4*a^2*b^8 + b^10)*d^2*sqr
t((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) - sqrt(2)*(2*(9*a^5*b^3 - 6*a^3*b^5 + a*b^7)*d^3*sqrt(
(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)*cos(d*x + c) + (9*a^8*b^3 + 12*a^6*b^5 - 2*a^4*b^7 - 4*a^2*b^9 + b^11
)*d*cos(d*x + c))*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6 + (a^3 - 3*a*b^2)*d^2*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*
b^4 + b^6)/d^4))/(9*a^4*b^2 - 6*a^2*b^4 + b^6))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*((a^6 + 3
*a^4*b^2 + 3*a^2*b^4 + b^6)/d^4)^(1/4) + (9*a^11*b^2 + 21*a^9*b^4 + 10*a^7*b^6 - 6*a^5*b^8 - 3*a^3*b^10 + a*b^
12)*cos(d*x + c) + (9*a^10*b^3 + 21*a^8*b^5 + 10*a^6*b^7 - 6*a^4*b^9 - 3*a^2*b^11 + b^13)*sin(d*x + c))/((a^2
+ b^2)*cos(d*x + c))) + 8*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x
 + c)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d)

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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(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.20, size = 822, normalized size = 7.41 \[ \frac {2 b \sqrt {a +b \tan \left (d x +c \right )}}{d}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a}{4 d b}+\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}-\frac {b \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}+\frac {2 b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}\, a}{4 d b}-\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a^{2}}{4 d b}+\frac {b \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {2 b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*(a+b*tan(d*x+c))^(1/2)/d-1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a
^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a+1/4/d/b*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2
)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/d*b*ln(b*tan(d*x+c)+a+(
a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+2/d*b/(2*(a
^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*
a)^(1/2))*a-1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2)
)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)+1/4/d/b*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/
2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*a-1/4/d/b*ln((a+b*tan(d*x+c))
^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/4/d*b
*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*
a)^(1/2)-2/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(
2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/d*b/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a
+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive, negative or zero?

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mupad [B]  time = 5.43, size = 1099, normalized size = 9.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^(3/2),x)

[Out]

(2*b*(a + b*tan(c + d*x))^(1/2))/d - atan((b^6*(a + b*tan(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2) + (3
*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2)*32i)/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16i)/d + (32*a^3*b^5)/
d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (32*a*b^5*(a + b*tan(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2) +
 (3*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2))/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16i)/d + (32*a^3*b^5)/d
 + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (a^2*b^4*(a + b*tan(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2) + (
3*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2)*96i)/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16i)/d + (32*a^3*b^5)
/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d) + (96*a^3*b^3*(a + b*tan(c + d*x))^(1/2)*((b^3*1i)/(4*d^2) - a^3/(4*d^2
) + (3*a*b^2)/(4*d^2) - (a^2*b*3i)/(4*d^2))^(1/2))/((a^2*b^6*32i)/d - (16*a*b^7)/d - (b^8*16i)/d + (32*a^3*b^5
)/d + (a^4*b^4*48i)/d + (48*a^5*b^3)/d))*((3*a*b^2 - a^2*b*3i - a^3 + b^3*1i)/(4*d^2))^(1/2)*2i - atan((b^6*(a
 + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^2))^(1/2)*32i)/
((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/d) + (32*a*b^5
*(a + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^2))^(1/2))/(
(b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (a^2*b^4*(
a + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^2))^(1/2)*96i)
/((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/d) - (96*a^3*
b^3*(a + b*tan(c + d*x))^(1/2)*((3*a*b^2)/(4*d^2) - (b^3*1i)/(4*d^2) - a^3/(4*d^2) + (a^2*b*3i)/(4*d^2))^(1/2)
)/((b^8*16i)/d - (16*a*b^7)/d - (a^2*b^6*32i)/d + (32*a^3*b^5)/d - (a^4*b^4*48i)/d + (48*a^5*b^3)/d))*((3*a*b^
2 + a^2*b*3i - a^3 - b^3*1i)/(4*d^2))^(1/2)*2i

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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