Optimal. Leaf size=219 \[ -\frac {(i A-B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {(i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]
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Rubi [A]
time = 1.29, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3686, 3736,
6857, 65, 223, 212, 95, 211, 214} \begin {gather*} \frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}-\frac {(-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 (-b+i a)^{3/2}}-\frac {(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 (b+i a)^{3/2}}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 95
Rule 211
Rule 212
Rule 214
Rule 223
Rule 3686
Rule 3736
Rule 6857
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \int \frac {-\frac {1}{2} a (A b-a B)+\frac {1}{2} b (A b-a B) \tan (c+d x)+\frac {1}{2} \left (a^2+b^2\right ) B \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {-\frac {1}{2} a (A b-a B)+\frac {1}{2} b (A b-a B) x+\frac {1}{2} \left (a^2+b^2\right ) B x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {Subst}\left (\int \left (\frac {\left (a^2+b^2\right ) B}{2 \sqrt {x} \sqrt {a+b x}}-\frac {b (a A+b B)-b (A b-a B) x}{2 \sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \frac {b (a A+b B)-b (A b-a B) x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}+\frac {B \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{b d}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {\text {Subst}\left (\int \left (\frac {b (A b-a B)+i b (a A+b B)}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {-b (A b-a B)+i b (a A+b B)}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b \left (a^2+b^2\right ) d}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{b d}\\ &=\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {((i a+b) (A+i B)) \text {Subst}\left (\int \frac {1}{(i-x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2+b^2\right ) d}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b d}-\frac {(i A+B) \text {Subst}\left (\int \frac {1}{\sqrt {x} (i+x) \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b) d}\\ &=\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}-\frac {((i a+b) (A+i B)) \text {Subst}\left (\int \frac {1}{i-(a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right ) d}-\frac {(i A+B) \text {Subst}\left (\int \frac {1}{i-(-a+i b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b) d}\\ &=-\frac {(i A-B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {2 B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{3/2} d}-\frac {(i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}+\frac {2 a (A b-a B) \sqrt {\tan (c+d x)}}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in
optimal.
time = 39.01, size = 177751, normalized size = 811.65 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 4.06, size = 1561438, normalized size = 7129.85 \[\text {output too large to display}\]
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 ) \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, 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 = -0.00 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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