Optimal. Leaf size=199 \[ \frac {\sqrt {i a-b} (i A-B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {i a+b} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}} \]
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Rubi [A]
time = 0.50, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3689, 3730,
3697, 3696, 95, 209, 212} \begin {gather*} \frac {\sqrt {-b+i a} (-B+i A) \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 (3 a B+A b) \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {\sqrt {b+i a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 95
Rule 209
Rule 212
Rule 3689
Rule 3696
Rule 3697
Rule 3730
Rubi steps
\begin {align*} \int \frac {\sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int \frac {\frac {1}{2} (-A b-3 a B)+\frac {3}{2} (a A-b B) \tan (c+d x)+A b \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}+\frac {4 \int \frac {-\frac {3}{4} a (a A-b B)-\frac {3}{4} a (A b+a B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {1}{2} ((a-i b) (A-i B)) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} ((a+i b) (A+i B)) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {((a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {((a+i b) (A+i B)) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}-\frac {((a-i b) (A-i B)) \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {((a+i b) (A+i B)) \text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=\frac {\sqrt {i a-b} (i A-B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {i a+b} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 A \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+3 a B) \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.98, size = 194, normalized size = 0.97 \begin {gather*} \frac {-3 \sqrt [4]{-1} \sqrt {-a+i b} (i A+B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 (-1)^{3/4} \sqrt {a+i b} (A+i B) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {2 \sqrt {a+b \tan (c+d x)} (a A+(A b+3 a B) \tan (c+d x))}{a \tan ^{\frac {3}{2}}(c+d x)}}{3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.48, size = 2181119, normalized size = 10960.40 \[\text {output too large to display}\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 ) \sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 = -0.01 \begin {gather*} \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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