\(\int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\) [802]

Optimal result

Integrand size = 45, antiderivative size = 155 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(2 i A-5 B) (a+i a \tan (e+f x))^{3/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(2 i A-5 B) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

-1/7*(I*A+B)*(a+I*a*tan(f*x+e))^(3/2)/f/(c-I*c*tan(f*x+e))^(7/2)-1/35*(2*I*A-5*B)*(a+I*a*tan(f*x+e))^(3/2)/c/f
/(c-I*c*tan(f*x+e))^(5/2)-1/105*(2*I*A-5*B)*(a+I*a*tan(f*x+e))^(3/2)/c^2/f/(c-I*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {(-5 B+2 i A) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {(-5 B+2 i A) (a+i a \tan (e+f x))^{3/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

[In]

Int[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

-1/7*((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(f*(c - I*c*Tan[e + f*x])^(7/2)) - (((2*I)*A - 5*B)*(a + I*a*Tan
[e + f*x])^(3/2))/(35*c*f*(c - I*c*Tan[e + f*x])^(5/2)) - (((2*I)*A - 5*B)*(a + I*a*Tan[e + f*x])^(3/2))/(105*
c^2*f*(c - I*c*Tan[e + f*x])^(3/2))

Rule 37

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 3669

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

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {\sqrt {a+i a x} (A+B x)}{(c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {(a (2 A+5 i B)) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{7 f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(2 i A-5 B) (a+i a \tan (e+f x))^{3/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}+\frac {(a (2 A+5 i B)) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{35 c f} \\ & = -\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(2 i A-5 B) (a+i a \tan (e+f x))^{3/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(2 i A-5 B) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.89 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {a^2 \sec ^4(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (21 A+5 (5 A+2 i B) \cos (2 (e+f x))+5 (-2 i A+5 B) \sin (2 (e+f x)))}{210 c^3 f (i+\tan (e+f x))^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

[In]

Integrate[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

-1/210*(a^2*Sec[e + f*x]^4*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*(21*A + 5*(5*A + (2*I)*B)*Cos[2*(e + f*x)]
+ 5*((-2*I)*A + 5*B)*Sin[2*(e + f*x)]))/(c^3*f*(I + Tan[e + f*x])^3*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Ta
n[e + f*x]])

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (5 B -25 i \tan \left (f x +e \right ) B -5 B \tan \left (f x +e \right )^{2}-23 i A -10 A \tan \left (f x +e \right )+2 i A \tan \left (f x +e \right )^{2}\right )}{105 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{5}}\)	\(113\)
default	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (5 B -25 i \tan \left (f x +e \right ) B -5 B \tan \left (f x +e \right )^{2}-23 i A -10 A \tan \left (f x +e \right )+2 i A \tan \left (f x +e \right )^{2}\right )}{105 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{5}}\)	\(113\)
risch	\(-\frac {a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (15 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+15 B \,{\mathrm e}^{6 i \left (f x +e \right )}+42 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+35 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-35 B \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{420 c^{3} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(117\)
parts	\(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-23+10 i \tan \left (f x +e \right )+2 \tan \left (f x +e \right )^{2}\right )}{105 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{5}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-1+5 i \tan \left (f x +e \right )+\tan \left (f x +e \right )^{2}\right )}{21 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{5}}\)	\(167\)

[In]

int((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/105*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a/c^4*(1+tan(f*x+e)^2)*(5*B-25*I*B*tan(f*x+e)
-5*B*tan(f*x+e)^2-23*I*A-10*A*tan(f*x+e)+2*I*A*tan(f*x+e)^2)/(I+tan(f*x+e))^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.75 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {{\left (15 \, {\left (i \, A + B\right )} a e^{\left (9 i \, f x + 9 i \, e\right )} + 3 \, {\left (19 i \, A + 5 \, B\right )} a e^{\left (7 i \, f x + 7 i \, e\right )} + 7 \, {\left (11 i \, A - 5 \, B\right )} a e^{\left (5 i \, f x + 5 i \, e\right )} + 35 \, {\left (i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{420 \, c^{4} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/420*(15*(I*A + B)*a*e^(9*I*f*x + 9*I*e) + 3*(19*I*A + 5*B)*a*e^(7*I*f*x + 7*I*e) + 7*(11*I*A - 5*B)*a*e^(5*
I*f*x + 5*I*e) + 35*(I*A - B)*a*e^(3*I*f*x + 3*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*
I*e) + 1))/(c^4*f)

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate((a+I*a*tan(f*x+e))**(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(7/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**(3/2)*(A + B*tan(e + f*x))/(-I*c*(tan(e + f*x) + I))**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.20 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\frac {{\left (15 \, {\left (-i \, A - B\right )} a \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 42 i \, A a \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 35 \, {\left (-i \, A + B\right )} a \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 15 \, {\left (A - i \, B\right )} a \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 42 \, A a \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 35 \, {\left (A + i \, B\right )} a \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{420 \, c^{\frac {7}{2}} f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

1/420*(15*(-I*A - B)*a*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 42*I*A*a*cos(5/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) + 35*(-I*A + B)*a*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 15*(A -
I*B)*a*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 42*A*a*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*
f*x + 2*e))) + 35*(A + I*B)*a*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(7/2)*f)

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(3/2)/(-I*c*tan(f*x + e) + c)^(7/2), x)

Mupad [B] (verification not implemented)

Time = 10.30 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.39 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx=-\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (2\,e+2\,f\,x\right )\,35{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,42{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,15{}\mathrm {i}-35\,B\,\cos \left (2\,e+2\,f\,x\right )+15\,B\,\cos \left (6\,e+6\,f\,x\right )-35\,A\,\sin \left (2\,e+2\,f\,x\right )-42\,A\,\sin \left (4\,e+4\,f\,x\right )-15\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,35{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,15{}\mathrm {i}\right )}{420\,c^3\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(3/2))/(c - c*tan(e + f*x)*1i)^(7/2),x)

[Out]

-(a*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(2*e + 2*f*x)*35i +
A*cos(4*e + 4*f*x)*42i + A*cos(6*e + 6*f*x)*15i - 35*B*cos(2*e + 2*f*x) + 15*B*cos(6*e + 6*f*x) - 35*A*sin(2*e
+ 2*f*x) - 42*A*sin(4*e + 4*f*x) - 15*A*sin(6*e + 6*f*x) - B*sin(2*e + 2*f*x)*35i + B*sin(6*e + 6*f*x)*15i))/
(420*c^3*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))