3.1.84 \(\int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [84]

Optimal. Leaf size=168 \[ -\frac {i \sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {62 i (a+i a \tan (c+d x))^{3/2}}{105 a d} \]

[Out]

-I*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+8/35*I*(a+I*a*tan(d*x+c))^(1/2)/d-2
/35*I*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d+2/7*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3/d+62/105*I*(a+I*a*tan(
d*x+c))^(3/2)/a/d

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Rubi [A]
time = 0.23, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3641, 3678, 3673, 3608, 3561, 212} \begin {gather*} \frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {i \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]^4*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (((8*I)/35)*Sqrt[a + I*a*Tan[
c + d*x]])/d - (((2*I)/35)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d + (2*Tan[c + d*x]^3*Sqrt[a + I*a*Tan[c
+ d*x]])/(7*d) + (((62*I)/105)*(a + I*a*Tan[c + d*x])^(3/2))/(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]

Rule 3678

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

Rubi steps

\begin {align*} \int \tan ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx &=\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 \int \tan ^2(c+d x) \left (3 a+\frac {1}{2} i a \tan (c+d x)\right ) \sqrt {a+i a \tan (c+d x)} \, dx}{7 a}\\ &=-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {4 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-i a^2+\frac {31}{4} a^2 \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}-\frac {4 \int \sqrt {a+i a \tan (c+d x)} \left (-\frac {31 a^2}{4}-i a^2 \tan (c+d x)\right ) \, dx}{35 a^2}\\ &=\frac {8 i \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}+\int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {8 i \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}-\frac {(2 i a) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac {i \sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {2 i \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {62 i (a+i a \tan (c+d x))^{3/2}}{105 a d}\\ \end {align*}

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Mathematica [A]
time = 2.38, size = 105, normalized size = 0.62 \begin {gather*} \frac {\sqrt {a+i a \tan (c+d x)} \left (-i e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac {2}{105} \left (-46 (-i+\tan (c+d x))+3 \sec ^2(c+d x) (-i+5 \tan (c+d x))\right )\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(Sqrt[a + I*a*Tan[c + d*x]]*(((-I)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])/E^(I*(c + d*x)) + (
2*(-46*(-I + Tan[c + d*x]) + 3*Sec[c + d*x]^2*(-I + 5*Tan[c + d*x])))/105))/d

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Maple [A]
time = 0.26, size = 94, normalized size = 0.56

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{3}}\)	\(94\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{3}}\)	\(94\)

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)^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 120, normalized size = 0.71 \begin {gather*} \frac {i \, {\left (105 \, \sqrt {2} a^{\frac {11}{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 ) + 60 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 168 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3} + 280 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4}\right )}}{210 \, a^{5} 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)^4,x, algorithm="maxima")

[Out]

1/210*I*(105*sqrt(2)*a^(11/2)*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))) + 60*(I*a*tan(d*x + c) + a)^(7/2)*a^2 - 168*(I*a*tan(d*x + c) + a)^(5/2)*a^3 + 280*(I*a*ta
n(d*x + c) + a)^(3/2)*a^4)/(a^5*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 330 vs. \(2 (125) = 250\).
time = 0.49, size = 330, normalized size = 1.96 \begin {gather*} \frac {105 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 ) - 105 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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 ) - 16 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-23 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 28 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 35 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{210 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^4,x, algorithm="fricas")

[Out]

1/210*(105*sqrt(2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-a/d^2
)*log(4*((I*d*e^(2*I*d*x + 2*I*c) + I*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)) - 105*sqrt(2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)
*sqrt(-a/d^2)*log(4*((-I*d*e^(2*I*d*x + 2*I*c) - I*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)) - 16*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-23*I*e^(7*I*d*x + 7*I*c) - 28*I
*e^(5*I*d*x + 5*I*c) - 35*I*e^(3*I*d*x + 3*I*c)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*
I*d*x + 2*I*c) + d)

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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 ^{4}{\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)**4,x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)**4, 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+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^4,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 0.34, size = 109, normalized size = 0.65 \begin {gather*} \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,4{}\mathrm {i}}{5\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,2{}\mathrm {i}}{7\,a^3\,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}}}{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)^4*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

((a + a*tan(c + d*x)*1i)^(3/2)*4i)/(3*a*d) - ((a + a*tan(c + d*x)*1i)^(5/2)*4i)/(5*a^2*d) + ((a + a*tan(c + d*
x)*1i)^(7/2)*2i)/(7*a^3*d) - (2^(1/2)*(-a)^(1/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*
1i)/d

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