3.823 \(\int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=283 \[ \frac {2 a^{7/2} (6 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {a^3 (6 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^3 f}-\frac {2 a^2 (6 B+i A) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a (6 B+i A) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

2*a^(7/2)*(I*A+6*B)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))/c^(5/2)/f-a^3*(I
*A+6*B)*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c^3/f-2/3*a^2*(I*A+6*B)*(a+I*a*tan(f*x+e))^(3/2)/c^2
/f/(c-I*c*tan(f*x+e))^(1/2)-1/5*(I*A+B)*(a+I*a*tan(f*x+e))^(7/2)/f/(c-I*c*tan(f*x+e))^(5/2)+2/15*a*(I*A+6*B)*(
a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.35, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3588, 78, 47, 50, 63, 217, 203} \[ \frac {2 a^{7/2} (6 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {a^3 (6 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^3 f}-\frac {2 a^2 (6 B+i A) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a (6 B+i A) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(2*a^(7/2)*(I*A + 6*B)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(c^(
5/2)*f) - ((I*A + B)*(a + I*a*Tan[e + f*x])^(7/2))/(5*f*(c - I*c*Tan[e + f*x])^(5/2)) + (2*a*(I*A + 6*B)*(a +
I*a*Tan[e + f*x])^(5/2))/(15*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - (2*a^2*(I*A + 6*B)*(a + I*a*Tan[e + f*x])^(3/
2))/(3*c^2*f*Sqrt[c - I*c*Tan[e + f*x]]) - (a^3*(I*A + 6*B)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*
x]])/(c^3*f)

Rule 47

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

Rule 50

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

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] || LtQ
[p, n]))))

Rule 203

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3588

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*} \int \frac {(a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2} (A+B x)}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(a (A-6 i B)) \operatorname {Subst}\left (\int \frac {(a+i a x)^{5/2}}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a (i A+6 B) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}+\frac {\left (a^2 (A-6 i B)\right ) \operatorname {Subst}\left (\int \frac {(a+i a x)^{3/2}}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 c f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a (i A+6 B) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 (i A+6 B) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {\left (a^3 (A-6 i B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a (i A+6 B) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 (i A+6 B) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {a^3 (i A+6 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^3 f}-\frac {\left (a^4 (A-6 i B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a (i A+6 B) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 (i A+6 B) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {a^3 (i A+6 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^3 f}+\frac {\left (2 a^3 (i A+6 B)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{c^2 f}\\ &=-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a (i A+6 B) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 (i A+6 B) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {a^3 (i A+6 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^3 f}+\frac {\left (2 a^3 (i A+6 B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{c^2 f}\\ &=\frac {2 a^{7/2} (i A+6 B) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{7/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a (i A+6 B) (a+i a \tan (e+f x))^{5/2}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 (i A+6 B) (a+i a \tan (e+f x))^{3/2}}{3 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {a^3 (i A+6 B) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^3 f}\\ \end {align*}

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Mathematica [A]  time = 17.77, size = 528, normalized size = 1.87 \[ \frac {\cos ^4(e+f x) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) \sqrt {\sec (e+f x) (c \cos (e+f x)-i c \sin (e+f x))} \left ((A-6 i B) \cos (2 f x) \left (-\frac {2 \sin (e)}{3 c^3}-\frac {2 i \cos (e)}{3 c^3}\right )+(6 B+i A) \cos (4 f x) \left (\frac {2 \cos (e)}{15 c^3}+\frac {2 i \sin (e)}{15 c^3}\right )+(A-i B) \cos (6 f x) \left (\frac {\sin (3 e)}{5 c^3}-\frac {i \cos (3 e)}{5 c^3}\right )+(A-6 i B) \sin (2 f x) \left (\frac {2 \cos (e)}{3 c^3}-\frac {2 i \sin (e)}{3 c^3}\right )+(A-6 i B) \sin (4 f x) \left (-\frac {2 \cos (e)}{15 c^3}-\frac {2 i \sin (e)}{15 c^3}\right )+(A-i B) \sin (6 f x) \left (\frac {\cos (3 e)}{5 c^3}+\frac {i \sin (3 e)}{5 c^3}\right )+(A-6 i B) \left (-\frac {\sin (3 e)}{c^3}-\frac {i \cos (3 e)}{c^3}\right )\right )}{f (\cos (f x)+i \sin (f x))^3 (A \cos (e+f x)+B \sin (e+f x))}+\frac {2 (6 B+i A) \sqrt {e^{i f x}} e^{-i (4 e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \tan ^{-1}\left (e^{i (e+f x)}\right ) (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{c^2 f \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \sec ^{\frac {9}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{7/2} (A \cos (e+f x)+B \sin (e+f x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(2*(I*A + 6*B)*Sqrt[E^(I*f*x)]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*ArcTan[E^(I*(e + f*x))]*(a + I*
a*Tan[e + f*x])^(7/2)*(A + B*Tan[e + f*x]))/(c^2*E^(I*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f*Sec[e +
f*x]^(9/2)*(Cos[f*x] + I*Sin[f*x])^(7/2)*(A*Cos[e + f*x] + B*Sin[e + f*x])) + (Cos[e + f*x]^4*((A - (6*I)*B)*
Cos[2*f*x]*((((-2*I)/3)*Cos[e])/c^3 - (2*Sin[e])/(3*c^3)) + (I*A + 6*B)*Cos[4*f*x]*((2*Cos[e])/(15*c^3) + (((2
*I)/15)*Sin[e])/c^3) + (A - (6*I)*B)*(((-I)*Cos[3*e])/c^3 - Sin[3*e]/c^3) + (A - I*B)*Cos[6*f*x]*(((-1/5*I)*Co
s[3*e])/c^3 + Sin[3*e]/(5*c^3)) + (A - (6*I)*B)*((2*Cos[e])/(3*c^3) - (((2*I)/3)*Sin[e])/c^3)*Sin[2*f*x] + (A
- (6*I)*B)*((-2*Cos[e])/(15*c^3) - (((2*I)/15)*Sin[e])/c^3)*Sin[4*f*x] + (A - I*B)*(Cos[3*e]/(5*c^3) + ((I/5)*
Sin[3*e])/c^3)*Sin[6*f*x])*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x] - I*c*Sin[e + f*x])]*(a + I*a*Tan[e + f*x])^(7/2)
*(A + B*Tan[e + f*x]))/(f*(Cos[f*x] + I*Sin[f*x])^3*(A*Cos[e + f*x] + B*Sin[e + f*x]))

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fricas [B]  time = 2.90, size = 519, normalized size = 1.83 \[ -\frac {15 \, c^{3} \sqrt {\frac {{\left (4 \, A^{2} - 48 i \, A B - 144 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}} f \log \left (\frac {2 \, {\left ({\left ({\left (4 i \, A + 24 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (4 i \, A + 24 \, B\right )} a^{3} e^{\left (i \, f x + 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}} + {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {\frac {{\left (4 \, A^{2} - 48 i \, A B - 144 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}}\right )}}{{\left (i \, A + 6 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 6 \, B\right )} a^{3}}\right ) - 15 \, c^{3} \sqrt {\frac {{\left (4 \, A^{2} - 48 i \, A B - 144 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}} f \log \left (\frac {2 \, {\left ({\left ({\left (4 i \, A + 24 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (4 i \, A + 24 \, B\right )} a^{3} e^{\left (i \, f x + 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}} - {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {\frac {{\left (4 \, A^{2} - 48 i \, A B - 144 \, B^{2}\right )} a^{7}}{c^{5} f^{2}}}\right )}}{{\left (i \, A + 6 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, A + 6 \, B\right )} a^{3}}\right ) - 2 \, {\left ({\left (-12 i \, A - 12 \, B\right )} a^{3} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (8 i \, A + 48 \, B\right )} a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-40 i \, A - 240 \, B\right )} a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-60 i \, A - 360 \, B\right )} a^{3} e^{\left (i \, f x + 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}}}{60 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/60*(15*c^3*sqrt((4*A^2 - 48*I*A*B - 144*B^2)*a^7/(c^5*f^2))*f*log(2*(((4*I*A + 24*B)*a^3*e^(3*I*f*x + 3*I*e
) + (4*I*A + 24*B)*a^3*e^(I*f*x + 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^3*f*e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt((4*A^2 - 48*I*A*B - 144*B^2)*a^7/(c^5*f^2)))/((I*A + 6*B)*a^3*e^(2*I*
f*x + 2*I*e) + (I*A + 6*B)*a^3)) - 15*c^3*sqrt((4*A^2 - 48*I*A*B - 144*B^2)*a^7/(c^5*f^2))*f*log(2*(((4*I*A +
24*B)*a^3*e^(3*I*f*x + 3*I*e) + (4*I*A + 24*B)*a^3*e^(I*f*x + 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^3*f*e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt((4*A^2 - 48*I*A*B - 144*B^2)*a^7/(c^5*f^2
)))/((I*A + 6*B)*a^3*e^(2*I*f*x + 2*I*e) + (I*A + 6*B)*a^3)) - 2*((-12*I*A - 12*B)*a^3*e^(7*I*f*x + 7*I*e) + (
8*I*A + 48*B)*a^3*e^(5*I*f*x + 5*I*e) + (-40*I*A - 240*B)*a^3*e^(3*I*f*x + 3*I*e) + (-60*I*A - 360*B)*a^3*e^(I
*f*x + 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^3*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)

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maple [B]  time = 0.48, size = 833, normalized size = 2.94 \[ -\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, a^{3} \left (-60 i A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +60 i A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{3}\left (f x +e \right )\right ) a c +15 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{4}\left (f x +e \right )\right ) a c +246 i B \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-90 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{4}\left (f x +e \right )\right ) a c +360 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{3}\left (f x +e \right )\right ) a c +15 B \left (\tan ^{4}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-90 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c -94 i A \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-90 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c -46 A \left (\tan ^{3}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-474 i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+540 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c -360 B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -564 B \left (\tan ^{2}\left (f x +e \right )\right ) \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+26 i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+15 A \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +74 A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+141 B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{15 f \,c^{3} \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{4} \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a^3/c^3*(-60*I*A*ln((c*a*tan(f*x+e)+(c*a*(1+ta
n(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)*a*c+60*I*A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(
1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^3*a*c+15*A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/
2))/(c*a)^(1/2))*tan(f*x+e)^4*a*c+246*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)^3-90*I*B*ln((c*a
*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^4*a*c+360*B*ln((c*a*tan(f*x+e)+(
c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^3*a*c+15*B*tan(f*x+e)^4*(c*a*(1+tan(f*x+e)^2)
)^(1/2)*(c*a)^(1/2)-90*I*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*a*c-94*I*
A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)^2-90*A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(
c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^2*a*c-46*A*tan(f*x+e)^3*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-474*I*B*(
c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+540*I*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*
a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)^2*a*c-360*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*
a)^(1/2))*tan(f*x+e)*a*c-564*B*tan(f*x+e)^2*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)+26*I*A*(c*a*(1+tan(f*x+e)
^2))^(1/2)*(c*a)^(1/2)+15*A*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*a*c+74*A
*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+141*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+t
an(f*x+e)^2))^(1/2)/(tan(f*x+e)+I)^4/(c*a)^(1/2)

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maxima [B]  time = 1.27, size = 944, normalized size = 3.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

((450*(A - 6*I*B)*a^3*cos(2*f*x + 2*e) + (450*I*A + 2700*B)*a^3*sin(2*f*x + 2*e) + 450*(A - 6*I*B)*a^3)*arctan
2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
1) + (450*(A - 6*I*B)*a^3*cos(2*f*x + 2*e) + (450*I*A + 2700*B)*a^3*sin(2*f*x + 2*e) + 450*(A - 6*I*B)*a^3)*a
rctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) + 1) - (180*(A - I*B)*a^3*cos(2*f*x + 2*e) - (-180*I*A - 180*B)*a^3*sin(2*f*x + 2*e) + 180*(A - I*B)*a^3)
*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (300*(A - 3*I*B)*a^3*cos(2*f*x + 2*e) + (300*I*A + 900
*B)*a^3*sin(2*f*x + 2*e) + 300*(A - 3*I*B)*a^3)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (900*(A
- 5*I*B)*a^3*cos(2*f*x + 2*e) - (-900*I*A - 4500*B)*a^3*sin(2*f*x + 2*e) + 900*(A - 6*I*B)*a^3)*cos(1/2*arcta
n2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((225*I*A + 1350*B)*a^3*cos(2*f*x + 2*e) - 225*(A - 6*I*B)*a^3*sin(2
*f*x + 2*e) + (225*I*A + 1350*B)*a^3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arc
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((
-225*I*A - 1350*B)*a^3*cos(2*f*x + 2*e) + 225*(A - 6*I*B)*a^3*sin(2*f*x + 2*e) + (-225*I*A - 1350*B)*a^3)*log(
cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^
2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + ((-180*I*A - 180*B)*a^3*cos(2*f*x + 2*e) + 1
80*(A - I*B)*a^3*sin(2*f*x + 2*e) + (-180*I*A - 180*B)*a^3)*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)) + ((300*I*A + 900*B)*a^3*cos(2*f*x + 2*e) - 300*(A - 3*I*B)*a^3*sin(2*f*x + 2*e) + (300*I*A + 900*B)*a^3)*s
in(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + ((-900*I*A - 4500*B)*a^3*cos(2*f*x + 2*e) + 900*(A - 5*I
*B)*a^3*sin(2*f*x + 2*e) + (-900*I*A - 5400*B)*a^3)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt
(a)*sqrt(c)/((-450*I*c^3*cos(2*f*x + 2*e) + 450*c^3*sin(2*f*x + 2*e) - 450*I*c^3)*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(7/2))/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(7/2))/(c - c*tan(e + f*x)*1i)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(7/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

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