\(\int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [108]

Optimal result

Integrand size = 28, antiderivative size = 155 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-1/8*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+1/4*(I*A+B)/a^2/d/(a+I*a*
tan(d*x+c))^(1/2)+1/5*(I*A-B)/d/(a+I*a*tan(d*x+c))^(5/2)+1/6*(I*A+B)/a/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3607, 3560, 3561, 212} \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {B+i A}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {-B+i A}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {B+i A}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

[In]

Int[(A + B*Tan[c + d*x])/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

-1/4*((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(5/2)*d) + (I*A - B)/(5*d*(a
+ I*a*Tan[c + d*x])^(5/2)) + (I*A + B)/(6*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (I*A + B)/(4*a^2*d*Sqrt[a + I*a
*Tan[c + d*x]])

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 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(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] && LtQ[n
, 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 3607

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

Rubi steps \begin{align*} \text {integral}& = \frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(A-i B) \int \frac {1}{(a+i a \tan (c+d x))^{3/2}} \, dx}{2 a} \\ & = \frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(A-i B) \int \frac {1}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4 a^2} \\ & = \frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(i A+B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d} \\ & = -\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {i A-B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i A+B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i A+B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.89 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.49 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {6 i A-6 B-5 (A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (-i+\tan (c+d x))}{30 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Integrate[(A + B*Tan[c + d*x])/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((6*I)*A - 6*B - 5*(A - I*B)*Hypergeometric2F1[-3/2, 1, -1/2, (1 + I*Tan[c + d*x])/2]*(-I + Tan[c + d*x]))/(30
*d*(a + I*a*Tan[c + d*x])^(5/2))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {i B -A}{12 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{8 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )}{d}\)	\(123\)
default	\(\frac {2 i \left (-\frac {-\frac {A}{2}-\frac {i B}{2}}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {i B -A}{12 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B -A}{8 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )}{d}\)	\(123\)
parts	\(\frac {2 i A a \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}+\frac {1}{8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}+\frac {B \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}-\frac {1}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d}\)	\(190\)

[In]

int((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d*(-1/5*(-1/2*A-1/2*I*B)/(a+I*a*tan(d*x+c))^(5/2)-1/12/a*(-A+I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/8/a^2*(-A+I*B
)/(a+I*a*tan(d*x+c))^(1/2)-1/16*(A-I*B)/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (118) = 236\).

Time = 0.25 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.53 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} - {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (23 i \, A + 17 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (-17 i \, A - 8 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (-7 i \, A + 2 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(15*sqrt(1/2)*a^3*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)
*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))
+ (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(
a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2)) - (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A +
B)) + sqrt(2)*((23*I*A + 17*B)*e^(6*I*d*x + 6*I*c) - 2*(-17*I*A - 8*B)*e^(4*I*d*x + 4*I*c) - 2*(-7*I*A + 2*B)*
e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral((A + B*tan(c + d*x))/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \, {\left (\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (A - i \, B\right )} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (A - i \, B\right )} a + 12 \, {\left (A + i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a}\right )}}{240 \, a d} \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/240*I*(15*sqrt(2)*(A - I*B)*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)))/a^(3/2) + 4*(15*(I*a*tan(d*x + c) + a)^2*(A - I*B) + 10*(I*a*tan(d*x + c) + a)*(A - I*B)*a
+ 12*(A + I*B)*a^2)/((I*a*tan(d*x + c) + a)^(5/2)*a))/(a*d)

Giac [F]

\[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 7.61 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.32 \[ \int \frac {A+B \tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {A\,1{}\mathrm {i}}{5\,d}+\frac {A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{6\,a\,d}+\frac {A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,1{}\mathrm {i}}{4\,a^2\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {-\frac {B}{5}+\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6\,a}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \]

[In]

int((A + B*tan(c + d*x))/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

((A*1i)/(5*d) + (A*(a + a*tan(c + d*x)*1i)*1i)/(6*a*d) + (A*(a + a*tan(c + d*x)*1i)^2*1i)/(4*a^2*d))/(a + a*ta
n(c + d*x)*1i)^(5/2) + ((B*(a + a*tan(c + d*x)*1i))/(6*a) - B/5 + (B*(a + a*tan(c + d*x)*1i)^2)/(4*a^2))/(d*(a
+ a*tan(c + d*x)*1i)^(5/2)) + (2^(1/2)*A*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/(8*
(-a)^(5/2)*d) - (2^(1/2)*B*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/(8*a^(5/2)*d)