3.6.12 \(\int \tan ^3(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\) [512]

Optimal. Leaf size=181 \[ \frac {(a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d} \]

[Out]

(a-I*b)^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+I*b)^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+
I*b)^(1/2))/d-2*a*(a+b*tan(d*x+c))^(1/2)/d-2/3*(a+b*tan(d*x+c))^(3/2)/d-4/35*a*(a+b*tan(d*x+c))^(5/2)/b^2/d+2/
7*tan(d*x+c)*(a+b*tan(d*x+c))^(5/2)/b/d

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Rubi [A]
time = 0.26, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3647, 3711, 12, 3609, 3620, 3618, 65, 214} \begin {gather*} -\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{d}+\frac {(a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a - I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + ((a + I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[
c + d*x]]/Sqrt[a + I*b]])/d - (2*a*Sqrt[a + b*Tan[c + d*x]])/d - (2*(a + b*Tan[c + d*x])^(3/2))/(3*d) - (4*a*(
a + b*Tan[c + d*x])^(5/2))/(35*b^2*d) + (2*Tan[c + d*x]*(a + b*Tan[c + d*x])^(5/2))/(7*b*d)

Rule 12

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

Rule 65

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 214

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

Rule 3609

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3647

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 3711

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]

Rubi steps

\begin {align*} \int \tan ^3(c+d x) (a+b \tan (c+d x))^{3/2} \, dx &=\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {2 \int (a+b \tan (c+d x))^{3/2} \left (-a-\frac {7}{2} b \tan (c+d x)-a \tan ^2(c+d x)\right ) \, dx}{7 b}\\ &=-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {2 \int -\frac {7}{2} b \tan (c+d x) (a+b \tan (c+d x))^{3/2} \, dx}{7 b}\\ &=-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\int \tan (c+d x) (a+b \tan (c+d x))^{3/2} \, dx\\ &=-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\int (-b+a \tan (c+d x)) \sqrt {a+b \tan (c+d x)} \, dx\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\int \frac {-2 a b+\left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}+\frac {1}{2} \left (i (a-i b)^2\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} \left (i (a+i b)^2\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {(a-i b)^2 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a-i b x}} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac {(a+i b)^2 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {a+i b x}} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}-\frac {\left (i (a-i b)^2\right ) \text {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 {\left (i (a+i b)^2\right ) \text {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)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {4 a (a+b \tan (c+d x))^{5/2}}{35 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{5/2}}{7 b d}\\ \end {align*}

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Mathematica [A]
time = 1.68, size = 170, normalized size = 0.94 \begin {gather*} \frac {(a-i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-2 a \left (3 a^2+70 b^2\right )+b \left (3 a^2-35 b^2\right ) \tan (c+d x)+24 a b^2 \tan ^2(c+d x)+15 b^3 \tan ^3(c+d x)\right )}{105 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a - I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + ((a + I*b)^(3/2)*ArcTanh[Sqrt[a + b*Tan[
c + d*x]]/Sqrt[a + I*b]])/d + (2*Sqrt[a + b*Tan[c + d*x]]*(-2*a*(3*a^2 + 70*b^2) + b*(3*a^2 - 35*b^2)*Tan[c +
d*x] + 24*a*b^2*Tan[c + d*x]^2 + 15*b^3*Tan[c + d*x]^3))/(105*b^2*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(587\) vs. \(2(151)=302\).
time = 0.14, size = 588, normalized size = 3.25

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 b^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \,b^{2} \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (2 \sqrt {a^{2}+b^{2}}\, a -\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \ln \left (-b \tan \left (d x +c \right )-a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (-2 \sqrt {a^{2}+b^{2}}\, a +\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {-2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{b^{2} d}\)	\(588\)
default	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 b^{2} \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \,b^{2} \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (2 \sqrt {a^{2}+b^{2}}\, a -\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \ln \left (-b \tan \left (d x +c \right )-a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (-2 \sqrt {a^{2}+b^{2}}\, a +\frac {\left (-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}+2 \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a \right ) \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{2}\right ) \arctan \left (\frac {-2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{b^{2} d}\)	\(588\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/b^2*(1/7*(a+b*tan(d*x+c))^(7/2)-1/5*a*(a+b*tan(d*x+c))^(5/2)-1/3*b^2*(a+b*tan(d*x+c))^(3/2)-a*b^2*(a+b*tan
(d*x+c))^(1/2)+b^2*(1/8*(-(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a)*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))+1/2*(2*(a^2+b^2)^(1/2)*a-
1/2*(-(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a)*(2*(a^2+b^2)^(1/2)+2*a)
^(1/2))/(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))-1/8*(-(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a)*
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))+1/2*(-2*(a^2+b^2)^(1/
2)*a+1/2*(-(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)+2*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a)*(2*(a^2+b^2)^(1/2)
+2*a)^(1/2))/(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))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 4373 vs. \(2 (147) = 294\).
time = 2.36, size = 4373, normalized size = 24.16 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(420*sqrt(2)*b^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) + sq
rt(2)*((3*a^5 + 2*a^3*b^2 - a*b^4)*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) + (3*a^8 + 2*a^6*b^2 - 4*a^4*b^4 - 2*a^2*b^6 + b^8)*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)*(a*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) + (a^4 - b^4)*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 + 12*a^6*b^2 - 2*a^4*b^4 - 4*a^2*b^6 + b^8)*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)*((9*a^6 - 15*a^4*b^2 + 7*a^2*b^4 - b^6)*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^9 + 12*a^7*b^2 - 2*a^5*b^4 - 4*a^3*b^6 + a*b^8)*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))*s
qrt((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
+ 21*a^9*b^2 + 10*a^7*b^4 - 6*a^5*b^6 - 3*a^3*b^8 + a*b^10)*cos(d*x + c) + (9*a^10*b + 21*a^8*b^3 + 10*a^6*b^5
- 6*a^4*b^7 - 3*a^2*b^9 + b^11)*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^1
6))*cos(d*x + c)^3 + 420*sqrt(2)*b^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^1
3 + 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^5 + 2*a^3*b^2 - a*b^4)*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) + (3*a^8 + 2*a^6*b^2 - 4*a^4*b^4 - 2*a^2*b^6 + b^8)*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)*(a*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) + (a^4 - b^4)*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 + 12*a^6*b^2 - 2*a^4*b^4 - 4*a^2*b^6 + b^8)*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)*((9*a^6 - 15*a^4*b^2 + 7*a^2*b^4 - b^6)*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^9 + 12*a^7*b^2 - 2*a^5*b^4 - 4*a^3*b^6 + a*b^8)*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 + 21*a^9*b^2 + 10*a^7*b^4 - 6*a^5*b^6 - 3*a^3*b^8 + a*b^10)*cos(d*x + c) + (9*a^10*b + 21*a^8*b
^3 + 10*a^6*b^5 - 6*a^4*b^7 - 3*a^2*b^9 + b^11)*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))*cos(d*x + c)^3 + 105*sqrt(2)*((a^3*b^2 - 3*a*b^4)*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6)/d^4)*cos(d*x + c)^3 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*cos(d*x + c)^3)*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 + 12*a^6*b^2 - 2*a^4*b^4 - 4*a^2*b^6 + b^8)*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)*((9*a^6 - 15*a^4*b^2 + 7*a^2*b^4 - b^6)
*d^3*sqrt((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)/d...

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 20.21, size = 1229, normalized size = 6.79 \begin {gather*} \sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\left (2\,a\,\left (\frac {2\,a^2}{b^2\,d}-\frac {2\,\left (a^2+b^2\right )}{b^2\,d}\right )-\frac {2\,a^3}{b^2\,d}+\frac {2\,a\,\left (a^2+b^2\right )}{b^2\,d}\right )+\left (\frac {2\,a^2}{3\,b^2\,d}-\frac {2\,\left (a^2+b^2\right )}{3\,b^2\,d}\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}+\frac {2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{7/2}}{7\,b^2\,d}-\frac {2\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^2\,d}+\mathrm {atan}\left (\frac {b^6\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}-\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}+\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}\,32{}\mathrm {i}}{\frac {16\,b^8}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {32\,a^2\,b^6}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}-\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}-\frac {32\,a\,b^5\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}-\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}+\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}}{\frac {16\,b^8}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {32\,a^2\,b^6}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}-\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}-\frac {a^2\,b^4\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}-\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}+\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}\,96{}\mathrm {i}}{\frac {16\,b^8}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {32\,a^2\,b^6}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}-\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}+\frac {96\,a^3\,b^3\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}-\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}+\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}}{\frac {16\,b^8}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {32\,a^2\,b^6}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}-\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}\right )\,\sqrt {-\frac {-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {b^6\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}+\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}-\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}\,32{}\mathrm {i}}{\frac {32\,a^2\,b^6}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {16\,b^8}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}+\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}+\frac {32\,a\,b^5\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}+\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}-\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}}{\frac {32\,a^2\,b^6}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {16\,b^8}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}+\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}-\frac {a^2\,b^4\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}+\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}-\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}\,96{}\mathrm {i}}{\frac {32\,a^2\,b^6}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {16\,b^8}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}+\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}-\frac {96\,a^3\,b^3\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {a^3}{4\,d^2}+\frac {b^3\,1{}\mathrm {i}}{4\,d^2}-\frac {3\,a\,b^2}{4\,d^2}-\frac {a^2\,b\,3{}\mathrm {i}}{4\,d^2}}}{\frac {32\,a^2\,b^6}{d}-\frac {a\,b^7\,16{}\mathrm {i}}{d}-\frac {16\,b^8}{d}+\frac {a^3\,b^5\,32{}\mathrm {i}}{d}+\frac {48\,a^4\,b^4}{d}+\frac {a^5\,b^3\,48{}\mathrm {i}}{d}}\right )\,\sqrt {-\frac {-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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