3.8.94 \(\int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\) [794]

Optimal. Leaf size=208 \[ -\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^3 f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

-2/105*(3*I*A-4*B)*(a+I*a*tan(f*x+e))^(1/2)/c^3/f/(c-I*c*tan(f*x+e))^(1/2)-1/7*(I*A+B)*(a+I*a*tan(f*x+e))^(1/2
)/f/(c-I*c*tan(f*x+e))^(7/2)-1/35*(3*I*A-4*B)*(a+I*a*tan(f*x+e))^(1/2)/c/f/(c-I*c*tan(f*x+e))^(5/2)-2/105*(3*I
*A-4*B)*(a+I*a*tan(f*x+e))^(1/2)/c^2/f/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.18, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3669, 79, 47, 37} \begin {gather*} -\frac {2 (-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{105 c^3 f \sqrt {c-i c \tan (e+f x)}}-\frac {2 (-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {(-4 B+3 i A) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7*((I*A + B)*Sqrt[a + I*a*Tan[e + f*x]])/(f*(c - I*c*Tan[e + f*x])^(7/2)) - (((3*I)*A - 4*B)*Sqrt[a + I*a*T
an[e + f*x]])/(35*c*f*(c - I*c*Tan[e + f*x])^(5/2)) - (2*((3*I)*A - 4*B)*Sqrt[a + I*a*Tan[e + f*x]])/(105*c^2*
f*(c - I*c*Tan[e + f*x])^(3/2)) - (2*((3*I)*A - 4*B)*Sqrt[a + I*a*Tan[e + f*x]])/(105*c^3*f*Sqrt[c - I*c*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*} \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {A+B x}{\sqrt {a+i a x} (c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac {(a (3 A+4 i B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}+\frac {(2 a (3 A+4 i B)) \text {Subst}\left (\int \frac {1}{\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) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac {(2 a (3 A+4 i B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{105 c^2 f}\\ &=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac {(3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac {2 (3 i A-4 B) \sqrt {a+i a \tan (e+f x)}}{105 c^3 f \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 4.32, size = 136, normalized size = 0.65 \begin {gather*} \frac {\cos (e+f x) (7 (-12 i A+B) \cos (e+f x)+15 (-4 i A+3 B) \cos (3 (e+f x))-(3 A+4 i B) (7 \sin (e+f x)+15 \sin (3 (e+f x)))) (\cos (4 (e+f x))+i \sin (4 (e+f x))) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{420 c^4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(7*((-12*I)*A + B)*Cos[e + f*x] + 15*((-4*I)*A + 3*B)*Cos[3*(e + f*x)] - (3*A + (4*I)*B)*(7*Sin[
e + f*x] + 15*Sin[3*(e + f*x)]))*(Cos[4*(e + f*x)] + I*Sin[4*(e + f*x)])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I
*c*Tan[e + f*x]])/(420*c^4*f)

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Maple [A]
time = 0.40, size = 147, normalized size = 0.71

method	result	size
risch	\(-\frac {\sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (15 i A \,{\mathrm e}^{6 i \left (f x +e \right )}+15 B \,{\mathrm e}^{6 i \left (f x +e \right )}+63 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+21 B \,{\mathrm e}^{4 i \left (f x +e \right )}+105 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-35 B \,{\mathrm e}^{2 i \left (f x +e \right )}+105 i A -105 B \right )}{840 c^{3} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(135\)
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i B \left (\tan ^{4}\left (f x +e \right )\right )+30 i A \left (\tan ^{3}\left (f x +e \right )\right )+6 A \left (\tan ^{4}\left (f x +e \right )\right )-84 i B \left (\tan ^{2}\left (f x +e \right )\right )-40 B \left (\tan ^{3}\left (f x +e \right )\right )-75 i A \tan \left (f x +e \right )-63 A \left (\tan ^{2}\left (f x +e \right )\right )+13 i B +65 B \tan \left (f x +e \right )+36 A \right )}{105 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{5}}\)	\(147\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i B \left (\tan ^{4}\left (f x +e \right )\right )+30 i A \left (\tan ^{3}\left (f x +e \right )\right )+6 A \left (\tan ^{4}\left (f x +e \right )\right )-84 i B \left (\tan ^{2}\left (f x +e \right )\right )-40 B \left (\tan ^{3}\left (f x +e \right )\right )-75 i A \tan \left (f x +e \right )-63 A \left (\tan ^{2}\left (f x +e \right )\right )+13 i B +65 B \tan \left (f x +e \right )+36 A \right )}{105 f \,c^{4} \left (i+\tan \left (f x +e \right )\right )^{5}}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/c^4*(8*I*B*tan(f*x+e)^4+30*I*A*tan(f*x+e)^3+6*A
*tan(f*x+e)^4-84*I*B*tan(f*x+e)^2-40*B*tan(f*x+e)^3-75*I*A*tan(f*x+e)-63*A*tan(f*x+e)^2+13*I*B+65*B*tan(f*x+e)
+36*A)/(I+tan(f*x+e))^5

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 6.38, size = 138, normalized size = 0.66 \begin {gather*} -\frac {{\left (15 \, {\left (i \, A + B\right )} e^{\left (9 i \, f x + 9 i \, e\right )} + 6 \, {\left (13 i \, A + 6 \, B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} + 14 \, {\left (12 i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + 70 \, {\left (3 i \, A - 2 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + 105 \, {\left (i \, A - B\right )} 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}}}{840 \, c^{4} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-1/840*(15*(I*A + B)*e^(9*I*f*x + 9*I*e) + 6*(13*I*A + 6*B)*e^(7*I*f*x + 7*I*e) + 14*(12*I*A - B)*e^(5*I*f*x +
5*I*e) + 70*(3*I*A - 2*B)*e^(3*I*f*x + 3*I*e) + 105*(I*A - B)*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^4*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(7/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(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)*sqrt(I*a*tan(f*x + e) + a)/(-I*c*tan(f*x + e) + c)^(7/2), x)

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Mupad [B]
time = 10.80, size = 246, normalized size = 1.18 \begin {gather*} -\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}}\,\left (A\,105{}\mathrm {i}-105\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,105{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,63{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,15{}\mathrm {i}-35\,B\,\cos \left (2\,e+2\,f\,x\right )+21\,B\,\cos \left (4\,e+4\,f\,x\right )+15\,B\,\cos \left (6\,e+6\,f\,x\right )-105\,A\,\sin \left (2\,e+2\,f\,x\right )-63\,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 (4\,e+4\,f\,x\right )\,21{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,15{}\mathrm {i}\right )}{840\,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}}} \end {gather*}

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)^(1/2))/(c - c*tan(e + f*x)*1i)^(7/2),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)*(A*105i - 105*B + A*cos(2*e
+ 2*f*x)*105i + A*cos(4*e + 4*f*x)*63i + A*cos(6*e + 6*f*x)*15i - 35*B*cos(2*e + 2*f*x) + 21*B*cos(4*e + 4*f*x
) + 15*B*cos(6*e + 6*f*x) - 105*A*sin(2*e + 2*f*x) - 63*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(4*e + 4*f*x)*21i + B*sin(6*e + 6*f*x)*15i))/(840*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))

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