\(\int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [847]

Optimal result

Integrand size = 45, antiderivative size = 157 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 i A+3 B) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 157, 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} \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {(3 B+2 i A) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(3 B+2 i A) \sqrt {c-i c \tan (e+f x)}}{15 a f (a+i a \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 3.53 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.63 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {c-i c \tan (e+f x)} \left (-7 i A-3 B+(6 A-9 i B) \tan (e+f x)+(2 i A+3 B) \tan ^2(e+f x)\right )}{15 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \]

[In]

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

[Out]

(Sqrt[c - I*c*Tan[e + f*x]]*((-7*I)*A - 3*B + (6*A - (9*I)*B)*Tan[e + f*x] + ((2*I)*A + 3*B)*Tan[e + f*x]^2))/
(15*a^2*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}+3 B \tan \left (f x +e \right )^{3}-13 i A \tan \left (f x +e \right )+8 A \tan \left (f x +e \right )^{2}+3 i B -12 B \tan \left (f x +e \right )-7 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\)	\(127\)
default	\(-\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \tan \left (f x +e \right )^{3}-12 i B \tan \left (f x +e \right )^{2}+3 B \tan \left (f x +e \right )^{3}-13 i A \tan \left (f x +e \right )+8 A \tan \left (f x +e \right )^{2}+3 i B -12 B \tan \left (f x +e \right )-7 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\)	\(127\)
parts	\(-\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (8 i \tan \left (f x +e \right )^{2}-2 \tan \left (f x +e \right )^{3}-7 i+13 \tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}+\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i \tan \left (f x +e \right )^{2}-\tan \left (f x +e \right )^{3}-i+4 \tan \left (f x +e \right )\right )}{5 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\)	\(173\)

[In]

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

[Out]

-1/15*I/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^3*(2*I*A*tan(f*x+e)^3-12*I*B*tan(f*x+e)^2+3
*B*tan(f*x+e)^3-13*I*A*tan(f*x+e)+8*A*tan(f*x+e)^2+3*I*B-12*B*tan(f*x+e)-7*A)/(I-tan(f*x+e))^4

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (15 \, {\left (-i \, A - B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 5 \, {\left (-5 i \, A - 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (13 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{60 \, a^{3} f} \]

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/60*(15*(-I*A - B)*e^(6*I*f*x + 6*I*e) + 5*(-5*I*A - 3*B)*e^(4*I*f*x + 4*I*e) - (13*I*A - 3*B)*e^(2*I*f*x +
2*I*e) - 3*I*A + 3*B)*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)

Sympy [F]

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

[In]

integrate((c-I*c*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

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

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.32 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,15{}\mathrm {i}+15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,25{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,13{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+15\,B\,\cos \left (2\,e+2\,f\,x\right )-3\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+25\,A\,\sin \left (2\,e+2\,f\,x\right )+13\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{120\,a^3\,f} \]

[In]

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

[Out]

(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*((c*(cos(2*e + 2*f*x) - sin(2
*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*15i + 15*B + A*cos(2*e + 2*f*x)*25i + A*cos(4*e + 4*f*x)
*13i + A*cos(6*e + 6*f*x)*3i + 15*B*cos(2*e + 2*f*x) - 3*B*cos(4*e + 4*f*x) - 3*B*cos(6*e + 6*f*x) + 25*A*sin(
2*e + 2*f*x) + 13*A*sin(4*e + 4*f*x) + 3*A*sin(6*e + 6*f*x) - B*sin(2*e + 2*f*x)*15i + B*sin(4*e + 4*f*x)*3i +
B*sin(6*e + 6*f*x)*3i))/(120*a^3*f)