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

Optimal. Leaf size=184 \[ \frac {(a+b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {(a-b) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a-b) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 a \sqrt {\tan (c+d x)}}{d}+\frac {2 b \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(a+b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {(a-b) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a-b) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 a \sqrt {\tan (c+d x)}}{d}+\frac {2 b \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a + b)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]
)/(Sqrt[2]*d) + ((a - b)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a - b)*Log[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*a*Sqrt[Tan[c + d*x]])/d + (2*b*Tan[c + d*x]^(3/
2))/(3*d)

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

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

Rule 1165

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

Rule 1168

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

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 3534

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

Rubi steps

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

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Mathematica [C]  time = 0.24, size = 94, normalized size = 0.51 \[ \frac {3 \sqrt [4]{-1} (a-i b) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} (3 a+b \tan (c+d x))+3 \sqrt [4]{-1} (a+i b) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-1)^(1/4)*(a - I*b)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 3*(-1)^(1/4)*(a + I*b)*ArcTanh[(-1)^(3/4)*Sqrt
[Tan[c + d*x]]] + 2*Sqrt[Tan[c + d*x]]*(3*a + b*Tan[c + d*x]))/(3*d)

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fricas [B]  time = 0.61, size = 2757, normalized size = 14.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*sqrt(2)*d^5*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*
b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4)*arctan(-((a^8 + 2*a^6*b^2 -
2*a^2*b^6 - b^8)*d^4*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - sqrt(2)*(a*d^7*sqrt
((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - (a^2*b + b^3)*d^5*sqrt((a^4 - 2*a^2*b^2 + b^
4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*s
qrt(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + sqrt(2)*((a^4*b - 2*
a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - (a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*cos(d*
x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*sq
rt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) + (a^8 - 2*a^4*b^4 + b^8)*sin(d*x + c))/cos(
d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) - sqrt(2)*((a^5 - a*b^4)*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*s
qrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*d^5*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4))*sq
rt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(sin(d*x
+ c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4))/(a^12 + 2*a^10*b^2 - a^8*b^4 - 4*a^6*b^6 - a^4*b^8 +
2*a^2*b^10 + b^12))*cos(d*x + c) + 12*sqrt(2)*d^5*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*
a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4)*
arctan(((a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*d^4*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)
/d^4) + sqrt(2)*(a*d^7*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - (a^2*b + b^3)*d^5
*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)
/(a^4 - 2*a^2*b^2 + b^4))*sqrt(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x
+ c) - sqrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - (a^7 - a^5*b^2
- a^3*b^4 + a*b^6)*d*cos(d*x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/
(a^4 - 2*a^2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) + (a^8 - 2*a^4*b^
4 + b^8)*sin(d*x + c))/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4) + sqrt(2)*((a^5 - a*b^4)*d^7*sqrt((a^
4 + 2*a^2*b^2 + b^4)/d^4)*sqrt((a^4 - 2*a^2*b^2 + b^4)/d^4) - (a^6*b + a^4*b^3 - a^2*b^5 - b^7)*d^5*sqrt((a^4
- 2*a^2*b^2 + b^4)/d^4))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a
^2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(3/4))/(a^12 + 2*a^10*b^2 - a^8*b
^4 - 4*a^6*b^6 - a^4*b^8 + 2*a^2*b^10 + b^12))*cos(d*x + c) + 3*sqrt(2)*(2*a*b*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4
)/d^4)*cos(d*x + c) - (a^4 + 2*a^2*b^2 + b^4)*d*cos(d*x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4
) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4)*log(((a^6 - a^4*b^2 -
a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) + sqrt(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqr
t((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - (a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*cos(d*x + c))*sqrt((2*a*b*d^
2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*
x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) + (a^8 - 2*a^4*b^4 + b^8)*sin(d*x + c))/cos(d*x + c)) - 3*sqrt(2)*
(2*a*b*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - (a^4 + 2*a^2*b^2 + b^4)*d*cos(d*x + c))*sqrt((2*a*
b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^2*b^2 + b^4))*((a^4 + 2*a^2*b^2 +
b^4)/d^4)^(1/4)*log(((a^6 - a^4*b^2 - a^2*b^4 + b^6)*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - sqrt
(2)*((a^4*b - 2*a^2*b^3 + b^5)*d^3*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4)*cos(d*x + c) - (a^7 - a^5*b^2 - a^3*b^4 +
a*b^6)*d*cos(d*x + c))*sqrt((2*a*b*d^2*sqrt((a^4 + 2*a^2*b^2 + b^4)/d^4) + a^4 + 2*a^2*b^2 + b^4)/(a^4 - 2*a^
2*b^2 + b^4))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^4 + 2*a^2*b^2 + b^4)/d^4)^(1/4) + (a^8 - 2*a^4*b^4 + b^8)*si
n(d*x + c))/cos(d*x + c)) - 8*(3*(a^5 + 2*a^3*b^2 + a*b^4)*cos(d*x + c) + (a^4*b + 2*a^2*b^3 + b^5)*sin(d*x +
c))*sqrt(sin(d*x + c)/cos(d*x + c)))/((a^4 + 2*a^2*b^2 + b^4)*d*cos(d*x + c))

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

[Out]

Timed out

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maple [A]  time = 0.09, size = 234, normalized size = 1.27 \[ \frac {2 b \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {2 a \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}-\frac {a \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {b \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {b \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {b \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 147, normalized size = 0.80 \[ -\frac {6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left (a - b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left (a - b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 8 \, b \tan \left (d x + c\right )^{\frac {3}{2}} - 24 \, a \sqrt {\tan \left (d x + c\right )}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*sqrt(2)*(a + b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 6*sqrt(2)*(a + b)*arctan(-1/2*
sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt(2)*(a - b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) +
1) - 3*sqrt(2)*(a - b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 8*b*tan(d*x + c)^(3/2) - 24*a*sqr
t(tan(d*x + c)))/d

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mupad [B]  time = 5.06, size = 114, normalized size = 0.62 \[ \frac {2\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*tan(c + d*x)^(1/2))/d + (2*b*tan(c + d*x)^(3/2))/(3*d) + ((-1)^(1/4)*a*atan((-1)^(1/4)*tan(c + d*x)^(1/2)
)*1i)/d + ((-1)^(1/4)*a*atanh((-1)^(1/4)*tan(c + d*x)^(1/2))*1i)/d - ((-1)^(1/4)*b*atan((-1)^(1/4)*tan(c + d*x
)^(1/2)))/d + ((-1)^(1/4)*b*atanh((-1)^(1/4)*tan(c + d*x)^(1/2)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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