3.1.85 \(\int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [85]

Optimal. Leaf size=127 \[ \frac {\sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{15 a d} \]

[Out]

arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d-8/5*(a+I*a*tan(d*x+c))^(1/2)/d+2/5*(a+
I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d-2/15*(a+I*a*tan(d*x+c))^(3/2)/a/d

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3641, 3673, 3608, 3561, 212} \begin {gather*} \frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{15 a d}-\frac {8 \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*Sqrt[a + I*a*Tan[c + d*x]])/(5*
d) + (2*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (2*(a + I*a*Tan[c + d*x])^(3/2))/(15*a*d)

Rule 212

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

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3608

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rule 3641

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[
a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

Rule 3673

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*d*((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*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx &=\frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 \int \tan (c+d x) \left (2 a+\frac {1}{2} i a \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)} \, dx}{5 a}\\ &=\frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{15 a d}-\frac {2 \int \sqrt {a+i a \tan (c+d x)} \left (-\frac {i a}{2}+2 a \tan (c+d x)\right ) \, dx}{5 a}\\ &=-\frac {8 \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{15 a d}+i \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {8 \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{15 a d}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=\frac {\sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {2 (a+i a \tan (c+d x))^{3/2}}{15 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.80, size = 95, normalized size = 0.75 \begin {gather*} \frac {\sec ^2(c+d x) \left (-10+\frac {30 \sinh ^{-1}\left (e^{i (c+d x)}\right ) \cos ^3(c+d x)}{\sqrt {1+e^{2 i (c+d x)}}}-16 \cos (2 (c+d x))-i \sin (2 (c+d x))\right ) \sqrt {a+i a \tan (c+d x)}}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(-10 + (30*ArcSinh[E^(I*(c + d*x))]*Cos[c + d*x]^3)/Sqrt[1 + E^((2*I)*(c + d*x))] - 16*Cos[2*(
c + d*x)] - I*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(15*d)

________________________________________________________________________________________

Maple [A]
time = 0.20, size = 92, normalized size = 0.72

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{2}}\)	\(92\)
default	\(-\frac {2 \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{2}}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 120, normalized size = 0.94 \begin {gather*} -\frac {15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 20 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 60 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}}{30 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(
d*x + c) + a))) + 12*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 20*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 60*sqrt(I*a*tan(
d*x + c) + a)*a^4)/(a^4*d)

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (98) = 196\).
time = 0.46, size = 285, normalized size = 2.24 \begin {gather*} \frac {15 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{d^{2}}} \log \left (-4 \, {\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {a}{d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (17 \, e^{\left (5 i \, d x + 5 i \, c\right )} + 20 \, e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, e^{\left (i \, d x + i \, c\right )}\right )}}{30 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log(4*((d*e^(2*I*d*x + 2*I*
c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(a/d^2) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 15*sqrt(2)*(d*e
^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/d^2)*log(-4*((d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2
*I*d*x + 2*I*c) + 1))*sqrt(a/d^2) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*
c) + 1))*(17*e^(5*I*d*x + 5*I*c) + 20*e^(3*I*d*x + 3*I*c) + 15*e^(I*d*x + I*c)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*
e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)**3, x)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Mupad [B]
time = 4.15, size = 100, normalized size = 0.79 \begin {gather*} -\frac {2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a\,d}-\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^2\,d}-\frac {\sqrt {2}\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

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

________________________________________________________________________________________