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

Optimal. Leaf size=284 \[ \frac {2 c^{7/2} (-6 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {c^3 (-6 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 c^2 (-6 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 c (-6 B+i A) (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 284, 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} \[ \frac {2 c^{7/2} (-6 B+i A) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{5/2} f}+\frac {c^3 (-6 B+i A) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a^3 f}+\frac {2 c^2 (-6 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 a^2 f \sqrt {a+i a \tan (e+f x)}}-\frac {2 c (-6 B+i A) (c-i c \tan (e+f x))^{5/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 14.25, size = 205, normalized size = 0.72 \[ \frac {2 \sqrt {2} c^2 e^{-4 i (e+f x)} \left (\frac {c}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (15 i (A+6 i B) e^{5 i (e+f x)} \left (1+e^{2 i (e+f x)}\right ) \tan ^{-1}\left (e^{i (e+f x)}\right )+i A \left (-2 e^{2 i (e+f x)}+10 e^{4 i (e+f x)}+15 e^{6 i (e+f x)}+3\right )-3 B \left (-4 e^{2 i (e+f x)}+20 e^{4 i (e+f x)}+30 e^{6 i (e+f x)}+1\right )\right )}{15 a^2 f \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*c^2*(c/(1 + E^((2*I)*(e + f*x))))^(3/2)*(I*A*(3 - 2*E^((2*I)*(e + f*x)) + 10*E^((4*I)*(e + f*x)) +
15*E^((6*I)*(e + f*x))) - 3*B*(1 - 4*E^((2*I)*(e + f*x)) + 20*E^((4*I)*(e + f*x)) + 30*E^((6*I)*(e + f*x))) +
(15*I)*(A + (6*I)*B)*E^((5*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))*ArcTan[E^(I*(e + f*x))]))/(15*a^2*E^((4*I)*
(e + f*x))*f*Sqrt[a + I*a*Tan[e + f*x]])

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fricas [B]  time = 0.71, size = 537, normalized size = 1.89 \[ \frac {{\left (15 \, a^{3} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}} f e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-4 i \, A + 24 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A + 24 \, B\right )} c^{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 (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}}\right )}}{{\left (-i \, A + 6 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 6 \, B\right )} c^{3}}\right ) - 15 \, a^{3} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}} f e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {2 \, {\left ({\left ({\left (-4 i \, A + 24 \, B\right )} c^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A + 24 \, B\right )} c^{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 (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{3} f\right )} \sqrt {\frac {{\left (4 \, A^{2} + 48 i \, A B - 144 \, B^{2}\right )} c^{7}}{a^{5} f^{2}}}\right )}}{{\left (-i \, A + 6 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A + 6 \, B\right )} c^{3}}\right ) + 2 \, {\left ({\left (60 i \, A - 360 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (40 i \, A - 240 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-8 i \, A + 48 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (12 i \, A - 12 \, B\right )} c^{3}\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}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{60 \, a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.43, size = 835, normalized size = 2.94 \[ \frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{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 ) 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}-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}-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 )+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}-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 +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 \,a^{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+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^(5/2),x)

[Out]

1/15/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^3/a^3*(-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)*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))*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)-94*I*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)^2-474*I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^
(1/2)*tan(f*x+e)+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)-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+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+ta
n(f*x+e)^2))^(1/2)/(-tan(f*x+e)+I)^4/(c*a)^(1/2)

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maxima [B]  time = 1.30, size = 1032, normalized size = 3.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(900*(A + 6*I*B)*c^3*cos(6*f*x + 6*e) + 600*(A + 6*I*B)*c^3*cos(4*f*x + 4*e) - 120*(A + 6*I*B)*c^3*cos(2*f*x +
2*e) - (-900*I*A + 5400*B)*c^3*sin(6*f*x + 6*e) - (-600*I*A + 3600*B)*c^3*sin(4*f*x + 4*e) - (120*I*A - 720*B
)*c^3*sin(2*f*x + 2*e) + 180*(A + I*B)*c^3 + (450*(A + 6*I*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 450*(A + 6*I*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-450*I*A + 2700*B)*c^3*
sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-450*I*A + 2700*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e))))*arctan2(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)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
450*(A + 6*I*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-450*I*A + 2700*B)*c^3*sin(7/2*ar
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-450*I*A + 2700*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f
*x + 2*e))))*arctan2(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) - ((-225*I*A + 1350*B)*c^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (-22
5*I*A + 1350*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 225*(A + 6*I*B)*c^3*sin(7/2*arctan2
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 225*(A + 6*I*B)*c^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)))*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) - ((225*I*A - 1350*B)*c^3*cos(7/2*arct
an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (225*I*A - 1350*B)*c^3*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) - 225*(A + 6*I*B)*c^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 225*(A + 6*I*B)*c^3*sin(
5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*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))*sqrt(a)*sqrt(c)/((-450*I*a^3*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 450*I*a^3*cos(5/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 450*a^3*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) +
450*a^3*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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