3.6.4 \(\int \tan ^3(c+d x) \sqrt {a+b \tan (c+d x)} \, dx\) [504]

Optimal. Leaf size=159 \[ \frac {\sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d}-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d} \]

[Out]

arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))*(a-I*b)^(1/2)/d+arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))*(a
+I*b)^(1/2)/d-2*(a+b*tan(d*x+c))^(1/2)/d-4/15*a*(a+b*tan(d*x+c))^(3/2)/b^2/d+2/5*tan(d*x+c)*(a+b*tan(d*x+c))^(
3/2)/b/d

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Rubi [A]
time = 0.23, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 11, 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))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{d}+\frac {\sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} \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*Sqrt[a + b*Tan[c + d*x]],x]

[Out]

(Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + (Sqrt[a + I*b]*ArcTanh[Sqrt[a + b*Tan[c +
d*x]]/Sqrt[a + I*b]])/d - (2*Sqrt[a + b*Tan[c + d*x]])/d - (4*a*(a + b*Tan[c + d*x])^(3/2))/(15*b^2*d) + (2*Ta
n[c + d*x]*(a + b*Tan[c + d*x])^(3/2))/(5*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) \sqrt {a+b \tan (c+d x)} \, dx &=\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {2 \int \sqrt {a+b \tan (c+d x)} \left (-a-\frac {5}{2} b \tan (c+d x)-a \tan ^2(c+d x)\right ) \, dx}{5 b}\\ &=-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {2 \int -\frac {5}{2} b \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx}{5 b}\\ &=-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\int \tan (c+d x) \sqrt {a+b \tan (c+d x)} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{d}-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\int \frac {-b+a \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{d}-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {1}{2} (-i a-b) \int \frac {1+i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (i a-b) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 \sqrt {a+b \tan (c+d x)}}{d}-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}-\frac {(a-i b) \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) \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 \sqrt {a+b \tan (c+d x)}}{d}-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}+\frac {(i a-b) \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 {(i a+b) \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 {\sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} \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)}}{d}-\frac {4 a (a+b \tan (c+d x))^{3/2}}{15 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{5 b d}\\ \end {align*}

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Mathematica [A]
time = 1.12, size = 140, normalized size = 0.88 \begin {gather*} \frac {15 \sqrt {a-i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+15 \sqrt {a+i b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-2 a^2-15 b^2+a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)\right )}{b^2}}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*Sqrt[a + b*Tan[c + d*x]],x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(369\) vs. \(2(133)=266\).
time = 0.14, size = 370, normalized size = 2.33

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 b^{2} \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (-b \tan \left (d x +c \right )-a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {-2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{b^{2} d}\)	\(370\)
default	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 a \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 b^{2} \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (-b \tan \left (d x +c \right )-a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-\sqrt {a^{2}+b^{2}}\right )}{8}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {-2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{2 \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{b^{2} d}\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

2/d/b^2*(1/5*(a+b*tan(d*x+c))^(5/2)-1/3*a*(a+b*tan(d*x+c))^(3/2)-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)*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*((a^2+b^2)^(1/2)-a)/(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)*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*(a-(a^2+b^2)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arcta
n((-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((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(d*x + c) + a)*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. 1879 vs. \(2 (129) = 258\).
time = 1.58, size = 1879, normalized size = 11.82 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*sqrt(2)*b^2*d^5*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)
^(3/4)*arctan(-((a^2 + b^2)*d^4*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) + (a^3 + a*b^2)*d^2*sqrt(b^2/d^4) + sqrt(2
)*(d^7*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) + a*d^5*sqrt(b^2/d^4))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d
*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(3/4) - sqrt(2)*(d^7*sqrt(b^2/
d^4)*sqrt((a^2 + b^2)/d^4) + a*d^5*sqrt(b^2/d^4))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(((
a^2 + b^2)*d^2*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 +
b^2)*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4)
- a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b
^2)*cos(d*x + c)))*((a^2 + b^2)/d^4)^(3/4))/(a^2*b^2 + b^4))*cos(d*x + c)^2 + 60*sqrt(2)*b^2*d^5*sqrt(-(a*d^2*
sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)^(3/4)*arctan(((a^2 + b^2)*d^4*sqrt(b^2
/d^4)*sqrt((a^2 + b^2)/d^4) + (a^3 + a*b^2)*d^2*sqrt(b^2/d^4) - sqrt(2)*(d^7*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^
4) + a*d^5*sqrt(b^2/d^4))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d
^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(3/4) + sqrt(2)*(d^7*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) + a*d^5*sqrt(
b^2/d^4))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*sqrt(((a^2 + b^2)*d^2*sqrt((a^2 + b^2)/d^4)*cos
(d*x + c) - sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 + b^2)*d*cos(d*x + c))*sqrt((a*cos(d*x +
c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4
) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*((a^2 + b^2)/d^4)^(3/
4))/(a^2*b^2 + b^4))*cos(d*x + c)^2 - 15*sqrt(2)*(a*b^2*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c)^2 + (a^2*b^2 +
b^4)*d*cos(d*x + c)^2)*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4)*log(((a^2
+ b^2)*d^2*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 + b^2
)*d*cos(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^
2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b^2)*
cos(d*x + c))) + 15*sqrt(2)*(a*b^2*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c)^2 + (a^2*b^2 + b^4)*d*cos(d*x + c)^2
)*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4)*log(((a^2 + b^2)*d^2*sqrt((a^2
+ b^2)/d^4)*cos(d*x + c) - sqrt(2)*(a*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^2 + b^2)*d*cos(d*x + c))*sqr
t((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(-(a*d^2*sqrt((a^2 + b^2)/d^4) - a^2 - b^2)/b^2)*((a^2 +
b^2)/d^4)^(1/4) + (a^3 + a*b^2)*cos(d*x + c) + (a^2*b + b^3)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c))) - 8*(3
*a^2*b^2 + 3*b^4 - 2*(a^4 + 10*a^2*b^2 + 9*b^4)*cos(d*x + c)^2 + (a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c))*sq
rt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c)))/((a^2*b^2 + b^4)*d*cos(d*x + c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Mupad [B]
time = 8.49, size = 354, normalized size = 2.23 \begin {gather*} \left (\frac {2\,a^2}{b^2\,d}-\frac {2\,\left (a^2+b^2\right )}{b^2\,d}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+\frac {2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^2\,d}-\frac {2\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^2\,d}+\mathrm {atan}\left (\frac {b^4\,\sqrt {\frac {a}{4\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}+\frac {32\,a\,b^3\,\sqrt {\frac {a}{4\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a-b\,1{}\mathrm {i}}{4\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {b^4\,\sqrt {\frac {a}{4\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}-\frac {32\,a\,b^3\,\sqrt {\frac {a}{4\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {b^5\,16{}\mathrm {i}}{d}+\frac {a^2\,b^3\,16{}\mathrm {i}}{d}}\right )\,\sqrt {\frac {a+b\,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))^(1/2),x)

[Out]

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

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