3.2.7 \(\int (a+i a \tan (c+d x))^{7/2} \, dx\) [107]

Optimal. Leaf size=130 \[ -\frac {8 i \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d} \]

[Out]

-8*I*a^(7/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d+8*I*a^3*(a+I*a*tan(d*x+c))^(1/2)/
d+4/3*I*a^2*(a+I*a*tan(d*x+c))^(3/2)/d+2/5*I*a*(a+I*a*tan(d*x+c))^(5/2)/d

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3559, 3561, 212} \begin {gather*} -\frac {8 i \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-8*I)*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((8*I)*a^3*Sqrt[a + I*a*Tan
[c + d*x]])/d + (((4*I)/3)*a^2*(a + I*a*Tan[c + d*x])^(3/2))/d + (((2*I)/5)*a*(a + I*a*Tan[c + d*x])^(5/2))/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 3559

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

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]

Rubi steps

\begin {align*} \int (a+i a \tan (c+d x))^{7/2} \, dx &=\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}+(2 a) \int (a+i a \tan (c+d x))^{5/2} \, dx\\ &=\frac {4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (4 a^2\right ) \int (a+i a \tan (c+d x))^{3/2} \, dx\\ &=\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (8 a^3\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {\left (16 i a^4\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac {8 i \sqrt {2} a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {8 i a^3 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {4 i a^2 (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 i a (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.85, size = 166, normalized size = 1.28 \begin {gather*} \frac {i \sqrt {2} a^3 e^{-\frac {1}{2} i (2 c+5 d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (\cos \left (\frac {3 d x}{2}\right )+i \sin \left (\frac {3 d x}{2}\right )\right ) \left (-120 \sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt {1+e^{2 i (c+d x)}} \sec ^3(c+d x) (35+38 \cos (2 (c+d x))+8 i \sin (2 (c+d x)))\right )}{15 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((I/15)*Sqrt[2]*a^3*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(Cos
[(3*d*x)/2] + I*Sin[(3*d*x)/2])*(-120*ArcSinh[E^(I*(c + d*x))] + Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^3*
(35 + 38*Cos[2*(c + d*x)] + (8*I)*Sin[2*(c + d*x)])))/(d*E^((I/2)*(2*c + 5*d*x)))

________________________________________________________________________________________

Maple [A]
time = 0.17, size = 92, normalized size = 0.71

method	result	size
derivativedivides	\(\frac {2 i a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(92\)
default	\(\frac {2 i a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d*a*(1/5*(a+I*a*tan(d*x+c))^(5/2)+2/3*a*(a+I*a*tan(d*x+c))^(3/2)+4*a^2*(a+I*a*tan(d*x+c))^(1/2)-4*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 = 0.50, size = 120, normalized size = 0.92 \begin {gather*} \frac {2 i \, {\left (30 \, \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 ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} + 10 \, {\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}\right )}}{15 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/15*I*(30*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))) + 3*(I*a*tan(d*x + c) + a)^(5/2)*a^2 + 10*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 60*sqrt(I*a*tan(
d*x + c) + a)*a^4)/(a*d)

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (95) = 190\).
time = 0.43, size = 321, normalized size = 2.47 \begin {gather*} \frac {4 \, {\left (15 \, \sqrt {2} \sqrt {-\frac {a^{7}}{d^{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 )} \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{7}}{d^{2}}} {\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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - 15 \, \sqrt {2} \sqrt {-\frac {a^{7}}{d^{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 )} \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{7}}{d^{2}}} {\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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - 2 \, \sqrt {2} {\left (-23 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 35 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{15 \, {\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))^(7/2),x, algorithm="fricas")

[Out]

4/15*(15*sqrt(2)*sqrt(-a^7/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^4*e^(I*d*x + I*
c) + sqrt(-a^7/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) -
15*sqrt(2)*sqrt(-a^7/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^4*e^(I*d*x + I*c) +
sqrt(-a^7/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) - 2*s
qrt(2)*(-23*I*a^3*e^(5*I*d*x + 5*I*c) - 35*I*a^3*e^(3*I*d*x + 3*I*c) - 15*I*a^3*e^(I*d*x + I*c))*sqrt(a/(e^(2*
I*d*x + 2*I*c) + 1)))/(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 \left (i a \tan {\left (c + d x \right )} + a\right )^{\frac {7}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Integral((I*a*tan(c + d*x) + a)**(7/2), 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))^(7/2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

Mupad [B]
time = 4.14, size = 107, normalized size = 0.82 \begin {gather*} \frac {a^3\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,8{}\mathrm {i}}{d}+\frac {a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,d}+\frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {\sqrt {2}\,{\left (-a\right )}^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,8{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(7/2),x)

[Out]

(a^3*(a + a*tan(c + d*x)*1i)^(1/2)*8i)/d + (a^2*(a + a*tan(c + d*x)*1i)^(3/2)*4i)/(3*d) + (a*(a + a*tan(c + d*
x)*1i)^(5/2)*2i)/(5*d) + (2^(1/2)*(-a)^(7/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*8i)/
d

________________________________________________________________________________________