Optimal. Leaf size=205 \[ \frac {(-1)^{3/4} (2 i A-B) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d} \]
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Rubi [A]
time = 0.45, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {3676, 3678,
3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {(-1)^{3/4} (-B+2 i A) \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 209
Rule 211
Rule 223
Rule 3625
Rule 3676
Rule 3678
Rule 3680
Rule 3682
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\frac {3}{2} a (i A-B)+a (A+2 i B) \tan (c+d x)\right ) \, dx}{a^2}\\ &=\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}-\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{2} a^2 (A+2 i B)+\frac {1}{2} a^2 (2 i A-B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a^3}\\ &=\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}-\frac {(A-i B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a}+\frac {(2 A+i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2}\\ &=\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {(2 A+i B) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {(a (i A+B)) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {(2 A+i B) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {(2 A+i B) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {\sqrt [4]{-1} (2 A+i B) \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (i A+B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(A+2 i B) \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}\\ \end {align*}
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Mathematica [A]
time = 3.05, size = 277, normalized size = 1.35 \begin {gather*} \frac {\left (\frac {\sqrt {2} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left ((i A+B) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+\sqrt {2} (-2 i A+B) \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \sqrt {\sec (c+d x)}}-2 ((A+2 i B) \cos (c+d x)-B \sin (c+d x)) \sqrt {\tan (c+d x)}\right ) (A+B \tan (c+d x))}{2 d (A \cos (c+d x)+B \sin (c+d x)) \sqrt {a+i a \tan (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1134 vs. \(2 (165 ) = 330\).
time = 0.15, size = 1135, normalized size = 5.54
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(1135\) |
default | \(\text {Expression too large to display}\) | \(1135\) |
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 807 vs. \(2 (155) = 310\).
time = 0.74, size = 807, normalized size = 3.94 \begin {gather*} -\frac {{\left (\sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {-i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - a d \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {104 \, {\left (2 \, \sqrt {2} {\left ({\left (2 i \, A - B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (2 i \, A - B\right )} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + {\left (3 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left ({\left (2 i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - B\right )}}\right ) + a d \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {104 \, {\left (2 \, \sqrt {2} {\left ({\left (2 i \, A - B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (2 i \, A - B\right )} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + {\left (-3 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {-4 i \, A^{2} + 4 \, A B + i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left ({\left (2 i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A - B\right )}}\right ) + 2 \, \sqrt {2} {\left ({\left (A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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